Mastering Projectile Motion: Practice Problems and Solutions

Expert reviewed • 22 November 2024 • 7 minute read

Revision of the key Concepts in Projectile Motion

Before diving into practice problems, let’s review the essential concepts:

Two-dimensional motion: Projectile motion combines horizontal and vertical components.
Constant horizontal velocity: Neglecting air resistance, the horizontal velocity remains constant.
Vertical motion: Affected by gravity, causing acceleration in the vertical direction.
Parabolic trajectory: The path of a projectile forms a symmetrical parabola.

Essential Equations

The following equations are crucial for solving projectile motion problems:

Horizontal position:

x_0=x+v_xt

Vertical position:

y = y_0 + v_{y0}t - \frac{1}{2}gt^2

Horizontal velocity:

v_x=v_0cos\theta

Vertical velocity:

v_y=v_0sin\theta-gt

Time of flight:

t_{flight} = \frac{2v_0 \sin(\theta)}{g}

Maximum height:

h_{max} = \frac{v_0^2 \sin^2(\theta)}{2g}

Horizontal range:

R = \frac{v_0^2 \sin(2\theta)}{g}

Where:

$v_0$ is the initial velocity
$\theta$ is the launch angle
$g$ is the acceleration due to gravity (approximately 9.8 m/s²)
$t$ is time

Practice Question 1

A ball is launched from ground level with an initial velocity of 20 m/s at an angle of 30° above the horizontal. Calculate:

The maximum height reached by the ball
The horizontal range of the ball

Maximum height:

Using the equation: $h_{max} = \frac{v_0^2 \sin^2(\theta)}{2g}$
$h_{max} = \frac{(20)^2 \sin^2(30°)}{2(9.8)} \\h_{max} = \frac{400 \times 0.25}{19.6} \\\approx 5.10 \text{ m}$
Horizontal range:

Using the equation: $R = \frac{v_0^2 \sin(2\theta)}{g}$
$R = \frac{(20)^2 \sin(60°)}{9.8 } \\R = \frac{400 \times 0.866}{9.8} \\\approx 35.3 \text{ m}$

$\therefore$ The ball reaches a maximum height of 5.10 m and travels a horizontal distance of 35.3 m.

Practice Question 2

A projectile is fired from a cliff 100 m above sea level with an initial velocity of 50 m/s at an angle of 45° above the horizontal. How long does it take for the projectile to hit the water?

We need to use the vertical position equation and solve for time:

y = y_0 + v_{y0}t - \frac{1}{2}gt^2

Where:

$y = 0$ is sea level
$y_0 = 100 m$ is the initial height
$v_{y0} = 50sin (45°) = 35.36 m/s$
$g = 9.8 m/s^2$

Substituting these values:

0 = 100 + 35.36t − 4.9t^2

This quadratic equation can be solved using the quadratic formula:

t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Where, a = − 4.9, b = 35.36, and c = 100

$a=-4.9$
$b=35.36$
$c=100$

Solving this gives us two solutions:

t_1 ≈ 8.54 s \quad \text{and}\quad t_2 ≈− 2.38 s

We discard the negative solution as time cannot be negative. Therefore, the projectile hits the water after approximately 8.54 seconds.

Practice Question 3

A baseball is hit at an angle of 40° above the horizontal. It reaches a maximum height of 30 m. What was the initial velocity of the baseball?

We can use the maximum height equation and solve for $v_0$ :

h_{max} = \frac{v_0^2 \sin^2(\theta)}{2g}

Rearranging for $v_0$ :

v_0 = \sqrt{\frac{2gh_{max}}{\sin^2(\theta)}}

Substituting the values:

v_0 = \sqrt{\frac{2(9.8 )(30 )}{\sin^2(40°)}} \\v_0 = \sqrt{\frac{588}{0.413}} \\\approx 37.7 \text{ m/s}

The initial velocity of the baseball was approximately 37.7 m/s.

Optimal Angle for Maximum Projectile Range

Return to Module 5: Advanced Mechanics