Table of contents

5. Projectile Motion

Intro to Projectile Motion: Horizontal Launch

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Bob throws a ball from his second story window to his friend Steve who is on the ground below. Assume that Bob throws the ball horizontally at $10 m / s$ , and Steve will catch the ball at a point $4.5 m$ below Bob. How far from the base of the building should Steve stand in order to catch the ball?

1.2 m

3.2 m

4.5 m

6.3 m

7.8 m

9.6 m

Verified step by step guidance

Identify the known values: the horizontal velocity of the ball is 10 m/s, and the vertical distance the ball falls is 4.5 m.

Use the kinematic equation for vertical motion to find the time it takes for the ball to fall 4.5 m. The equation is:

y = v

₀t + 12gt², where

v

₀ is the initial vertical velocity (0 m/s),

g

is the acceleration due to gravity (9.8 m/s²), and

y

is the vertical distance (4.5 m).

Since the initial vertical velocity is 0, the equation simplifies to:

y = \frac{1}{2} g t

². Solve for

t

to find the time it takes for the ball to fall 4.5 m.

Once you have the time, use the horizontal motion equation to find the horizontal distance:

x = v

₀t, where

v

₀ is the horizontal velocity (10 m/s) and

t

is the time calculated in the previous step.

Calculate the horizontal distance

x

to determine how far from the base of the building Steve should stand to catch the ball.

Master Introduction to Projectile Motion with a bite sized video explanation from Patrick

Problem sets built by lead tutors

Expert video explanations

Bob throws a ball from his second story window to his friend Steve who is on the ground below. Assume that Bob throws the ball horizontally at 10⁢m/s, and Steve will catch the ball at a point 4.5⁢m below Bob. How far from the base of the building should Steve stand in order to catch the ball?