Problems and Solutions to Practice Projectile Motion Physics

Zubair Sensei
Zubair Sensei
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Projectile Motion Problems And Solutions Explained Like I'm Your Tutor

We have five problems with their solutions for you to learn the physics of projectile motions. The objective of these problems is to help you achieve some competence listed below:
  • Find the x component and y component of an object’s initial velocity in a projectile motion
  • Find the maximum height of the projectile motion
  • Determine the time it took for an object to reach maximum height
  • Determine the horizontal position of an object when it reach maximum height
  • Find the time it took for an object to complete a projectile motion
  • Find the elevation angle of a projectile motion
  • Determining the velocity of an object in the air at a certain time or a certain position
  • Find the change of height after a period of time
Take your time to study. Prepare your blank sheet to try something on. I hope you could learn something new. Happy Learning.

1. Find the maximum height a projectile motion

A soccer player is going to take a free kick. The ball will be kicked with Cristiano Ronaldo’s knuckleball technique where the ball will experience a projectile-like motion without bending at all.
The initial velocity of the ball is 8.95 m/s and its angle of elevation is \alpha=47.85^{\circ}. Assume the ball undergoes a projectile motion. Determine the time it took for the ball to reach its maximum height. Also determine the maximum height of the ball. And determine the horizontal position where the ball is at its maximum height.
Information we have:
\begin{aligned} v_o=&8.95\;\text{m/s}\\ \alpha=&47.85^{\circ} \end{aligned}
What we need to find:
\begin{aligned} t_{peak}=&….?\\ h_{peak}=&….?\\ x_{peak}=&….? \end{aligned}
Method and Solution:
Normally in all kinds of projectile motion problems, the first thing to do is to find the x and y components of the object’s initial velocity. The method is as follows.
\begin{aligned} v_{ox}=&v_o \cos{\alpha}\\ =&8.95 \cos{(47.85^{\circ})}\\ =&6\;m/s \end{aligned}
\begin{aligned} v_{oy}=&v_o \sin{\alpha}\\ =&8.95 \sin{(47.85^{\circ})}\\ =&6.64\;m/s \end{aligned}
When the ball is at the peak of height, the vertical velocity of the ball ( v_y) is zero. We can use the equation for constant acceleration motion to find the time it took for the ball to reach the peak. The method is as below.
\begin{aligned} v_y=&v_{oy}+gt_{peak}\\ 0=&6.64+(-10)t\\ 10t_{peak}=&6.64\\ t_{peak}=&0.664\;\text{seconds} \end{aligned}
So it took approximately 0.664 seconds for the ball to reach the peak(maximum height of projectile motion). Next, we find the height of the object when it is at the top.
\begin{aligned} y=&v_{oy}t_{peak}+\frac{1}{2}g{t_{peak}}^2\\ =&(6.64)(0.664)+\frac{1}{2}(-10)(0.664^2)\\ =&2.2\;\text{m} \end{aligned}
So, the ball is 2.2 meters above the ground when it reached its peak. Next, we find the horizontal distance at which the ball is at the peak. The method is as follows.
\begin{aligned} x_{peak}=&v_{ox}\;t_{peak}\\ =&(6)(0.664)\\ \approx&\;4.0\;\text{m} \end{aligned}
The ball is horizontally 4 meters away from the initial kick position when it reached the peak of its projectile motion.
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2. Find the initial velocity of a projectile motion

A person strikes a golf ball so that it undergoes a projectile motion and hits the ground at a distance of 122.5 meters from its initial position. The ball is in the air for \frac{7}{2}\sqrt{2} seconds. Determine the initial velocity of the golf ball and its angle of elevation.
projectile motion of a golf ball problem with solution
Information we have:
\begin{aligned} x=&122.5\;m\\ t=&\frac{7}{2}\sqrt{2}\;\text{seconds}\\ \end{aligned}
What we need to find:
\begin{aligned} v_{o}=&…..?\\ \alpha=&…..? \end{aligned}
Method and Solution:
We must understand that horizontally the golf ball is moving with constant speed. And we can use this equation x=v_{\circ x}.t to get the x component of the initial velocity of the golf ball (v_{\circ x}).
\begin{aligned} x=&v_{ox}\;t\\ 122.5=&v_{ox}\;(\frac{7}{2}\sqrt{2})\\ v_{ox}=&\frac{35}{2}\sqrt{2}\;m/s \end{aligned}
Think deeply about the problem. The ball hits the ground after \frac{7}{2}\sqrt{2} seconds in the air. This means when y equals zero, t equals \frac{7}{2}\sqrt{2} seconds. We use the vertical motion equation to find the y component of the golf ball’s initial velocity. Look below to see how we would do it.
\begin{aligned} y=&v_{oy}\;t+\frac{1}{2}gt^2\\ 0=&v_{oy}\;(\frac{7}{2}\sqrt{2})+\frac{1}{2}(-10)(\frac{7}{2}\sqrt{2})^2\\ 0=&v_{oy}\;(\frac{7}{2}\sqrt{2})-(5)(\frac{49}{4})(2)\\ 122.5=&v_{oy}\;(\frac{7}{2}\sqrt{2})\\ v_{oy}=&\frac{35}{2}\sqrt{2}\;m/s \end{aligned}
So the y component of the golf ball’s initial velocity is \frac{35}{2}\sqrt{2}\;m/s. Next, we find the angle of elevation of the golf ball. The method is as follows.
\begin{aligned} \alpha=&\tan^{-1}\frac{v_{\circ y}}{v_{\circ x}}\\\\ =&\tan^{-1}\frac{\frac{35}{2}\sqrt{2}}{\frac{35}{2}\sqrt{2}}\\ =&45^{\circ} \end{aligned}
So the angle of elevation is 45^{\circ}.

3. Zero angle launch projectile motion

A car company is conducting research on the suspension of a car model they manufactured. A prototype car is moving 20 m/s towards the edge of a cliff that is 20 meters in height. The car then undergoes a projectile motion and falls to the bottom of the cliff. Calculate the time the car is in the air? Determine the horizontal distance traveled before the car touch the ground.
Soal gerak parabola uji mobil
Information we have:
\begin{aligned} v_{o}=&v_{ox}=20\;m/s\\ h=&20\;m \end{aligned}
What we need to find:
\begin{aligned} t=&…….?\\ x=&…….? \end{aligned}
Method and Solution:
For this type of problem, we must understand that the y component of the initial velocity is zero (v_{\circ y}=0) because the angle of elevation is zero (\alpha=0). This means the x component of the initial velocity of the car is (v_{\circ}=v_{\circ x}=20\; m/s).
First, let’s find out the duration of the car in the air. We could use the equation of vertical motion and that is y=v_{oy}\;t+\frac{1}{2}gt^2. And we must know that when the car hits the ground at the bottom of the cliff, its position is y = -20.
\begin{aligned} y=&v_{oy}\;t+\frac{1}{2}gt^2\\ -20=&(0)t+\frac{1}{2}(-10)t^2\\ t^2=&\frac{20}{5}\\ t=&\sqrt{4}\\ t=&2\;\text{seconds} \end{aligned}
Now we have found that the car is in the air for 2 seconds. With this information, we could find the horizontal distance the car travels before it touches the ground. The method is as follows.
\begin{aligned} x=&v_{ox}\;t\\ =&(20)(2)\\ =&40\;m \end{aligned}
So the car hits the ground 40 meters away from the cliff.

4. Projectile motion from a certain height

Soal rusa loncat dari tebing renewed
A roe deer jumps exactly at the edge of a cliff with an initial speed of 5 m/s and its angle of elevation \alpha=60^{\circ}. The deer is in the air for 2 seconds before it finally touches the water. Determine the horizontal location where the deer touches the water. Calculate the height of the cliff.
Information we have:
\begin{aligned} t=&2\;\text{seconds}\\ v_{o}=&5\;m/s\\ \alpha=&60^{\circ} \end{aligned}
What we need to find:
\begin{aligned} x=&…….?\\ h=&…….?\\ v_f=&…….? \end{aligned}
Method and Solution:
As usual, we look for the most important information, we need to find the x and y components of the initial velocity.
\begin{aligned} v_{ox}=&v_{\circ}\;\cos{\alpha}\\ =&5\cos{60^{\circ}}\\ =&2.5\;m/s \end{aligned}
\begin{aligned} v_{\circ y}=&v_{\circ}\;\sin{\alpha}\\ =&5\sin{60^{\circ}}\\ =&2.5\sqrt{3}\;m/s \end{aligned}
Since we already know the time it took for the deer to touch the water. We could find the horizontal distance travelled by the deer in the air. The method is as follows.
\begin{aligned} x=&v_{\circ x}\;t\\ =&(2.5)(2)\\ =&5\;m \end{aligned}
Next, we should find the vertical position (y) of the deer when t=2 seconds. The method is as below.
\begin{aligned} y=&v_{\circ y}\;t+\frac{1}{2}gt^2\\ =&(\frac{5}{2}\sqrt{3})(2)+\frac{1}{2}(-10)2^2\\ =&5\sqrt{3}-20\\ =&-11.34\;\text{m} \end{aligned}
We just found y=-11.34 meters. This means that when t=2 seconds the deer is 11.24 meters below the starting position. This also means the height of the cliff is 11.24 meters. So mathematically we can write as below.
\begin{aligned} h=\left | y \right |=&11.34\;\text{m} \end{aligned}
We can calculate the velocity of the deer when it splashed into the water using this equation.
v_f=\sqrt{{v_x}^2+{v_y}^2}
We already know v_x=v_{\circ x}=2.5\;m/s. But we don’t know v_y yet, so we have to calculate v_y(the vertical velocity of the deer when it touches the water or at t=2 seconds). The method is written down below.
\begin{aligned} v_y=&v_{\circ y}+gt\\ =&\frac{5}{2}\sqrt{3}+(-10)2\\ =&\frac{5}{2}\sqrt{3}-20\\ =&-15.67\;m/s\\ \end{aligned}
Now at this point, we already have all the information we need. We can calculate the speed of the deer when it hits the water (v_f).
\begin{aligned} v_f=&\sqrt{{v_x}^2+{v_y}^2}\\ =&\sqrt{{2.5}^2+({-15.67}^2)}\\ =&15.87\;m/s \end{aligned}
We have calculated the speed of the deer right when it plunges into the water and that is 15.86 m/s.

5. Basketball's projectile motion

gerak Parabola bola basket
A basketball player throws a basketball into the net. The initial velocity of the ball is \sqrt{70} m/s and its angle of elevation is \alpha where \sin{\alpha}=\frac{\sqrt{ 3}}{\sqrt{7}} and \cos{\alpha}=\frac{2}{\sqrt{7}}. The ball is thrown at 2 meters above the floor. The net is 3 meters high. Calculate the time the ball is in the air before it enters the net. Also, find the horizontal distance between the ring and the initial position of the ball. Find the velocity of the basketball as it enters the net and its angle from the horizontal.
Information we have:
\begin{aligned} v_{\circ}=&\sqrt{70}\;m/s\\ \sin{\alpha}=&\frac{\sqrt{3}}{\sqrt{7}}\\ \cos{\alpha}=&\frac{2}{\sqrt{7}}\\ h_1=&3\;m\\ h_2=&2\;m \end{aligned}
What we need to find:
\begin{aligned} t=&……….?\\ x=&……….?\\ v_{enter}=&……….?\\ \end{aligned}
Method and Solution:
Kita bisa menggunakan rumus cepat untuk mencari tinggi maksimum bola basket. Caranya di bawah ini.
\begin{aligned} h_{max}=&\frac{{v_{\circ}}^2 \sin^2{\alpha}}{2g}\\ =&\frac{{\sqrt{70}}^2. \left (\frac{\sqrt{3}}{\sqrt{7}} \right )^2}{2(10)}\\ =&1,5\;m \end{aligned}
Tapi perlu disadari bahwa ketinggian 1,5 meter yang barusan kita dapat adalah ketinggian dari posisi awal lemparan. Ketinggian maksimum yang bola basket dari lantai adalah 1,5 meter + ketinggian awal = 1,5 meter + 2 meter = 3,5 meter.
y=1 ketika bola masuk ke ring
When the ball enters the net, it is 1 meter above its initial position. This means y = 1. So we can use the y equation for vertical motion as follows.
\begin{aligned} y=&v_{\circ y}\;t+\frac{1}{2}gt^2\\ 1=&0.t+\frac{1}{2}(-10)t^2\\ 0=&-5t^2+\sqrt{30}t-1 \end{aligned}
Now we have a quadratic equation whose solution is:
\begin{aligned} t_1=&\frac{\sqrt{3}-1}{\sqrt{10}}\\ \text{and}\\ t_2=&\frac{\sqrt{3}+1}{\sqrt{10}} \end{aligned}
Ada 2 waktu ketika y=1
This means there are 2 specific times where the ball is at y=1 namely when t_1 and t_2. We can see in the picture, the ball goes into the ring at t_2. This means the ball is in the air for \frac{\sqrt{3}+1}{\sqrt{10}}\approx 0.86 seconds before entering the net.
Next we could calculate the distance between the initial position and the basketball net. We need to find x when t=t_2=\frac{\sqrt{3}+1}{\sqrt{10}}.
Mencari x gerak parabola
\begin{aligned} x=&v_x\;t\\ =&v_{\circ}\cos{\alpha}\;t\\ =&\sqrt{70}\times\frac{2}{\sqrt{7}}\times\frac{\sqrt{3}+1}{\sqrt{10}}\\ =&2+2\sqrt{3}\;m \end{aligned}
Translation results So the distance between the net and the basketball player is 2+2\sqrt{3}\;m or about 5.46 m.
Next, we would calculate the velocity of the ball when it enters the net. We could calculate it using the following equation.
\begin{aligned} v_{enter}=&\sqrt{{v_x}^2+{v_y}^2}\\ \end{aligned}
Mencari kecepatan bola basket ketika masuk ke ring
We need to find v_x first.
\begin{aligned} v_{x}=v_{\circ x}=&v_{\circ}\cos{\alpha}\\ =&\sqrt{70}\times\frac{2}{\sqrt{7}}\\ =&2\sqrt{10}\;m/s \end{aligned}
Next, we have to find v_y.
\begin{aligned} 2gy=&{v_y}^2-{v_{\circ y}}^2\\ 2(-10)(1)=&{v_y}^2-(\sqrt{30})^2\\ {v_y}^2=&10\\ v_y=&\sqrt{10}\;m/s \end{aligned}
So the velocity of the ball when it enters the net is as follows.
\begin{aligned} v_{enter}=&\sqrt{{v_x}^2+{v_y}^2}\\ =&\sqrt{({2\sqrt{10}})^2+({\sqrt{10}})^2}\\ =&\sqrt{30}\;m/s \end{aligned}
We could find the angle between the ball’s velocity and the horizontal when it enters the net. The method is as follows.
\begin{aligned} \theta=&\tan^{-1}\frac{v_y}{v_x}\\ =&\tan^{-1}\frac{\sqrt{10}}{2\sqrt{10}}\\ =&\tan^{-1}\frac{1}{2}\\ =&26.57^{\circ} \end{aligned}

Tips for working on projectile motion problems

In my opinion, there are 2 most important quantities to look for when working on projectile motion problems, they are:
  • The x and y components of the initial velocity of the projectile motion ( v_{\circ x} and v_{\circ y})
  • Time (t)
If the two quantities are known, we could find other quantities such as distance, altitude, and velocity.
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This Post Has One Comment

  1. Hello, the whole thing is going nicely here and ofcourse every one is sharing
    facts, that’s actually fine, keep up writing.

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