A ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec. Assuming that the air resistance can be ignored, how high does it go?

Question

Hint: The acceleration due to gravity is 32 ft per second squared.

Answer 1

The vertical position of the ball is given by:

y = y₀ + v_0yt + 0.5at²

Where,

y = vertical height of the ball from the ground after t seconds

y₀ = The height of the ball at launch (the ground) = 0ft

a = acceleration due to gravity = -32ft/sec²

t = time in seconds

Key Idea #1: Velocity is the first derivative of the position function.

The vertical velocity with respect to time is:

dy/dt = 0 + v_0y + 2(0.5)at

= v_0y + at

v = v_0y + at

where,

v = vertical velocity of the ball

v_0y = initial vertical velocity of the ball = 42 ft/sec

a = acceleration due to gravity = -32 ft/sec²

t = time in seconds

Key idea #2: At the time when the ball reaches its maximum height and the ball's path changes from going up to going down - at that instant in time, the velocity of the ball is zero.

By plugging in zero for velocity (v=0) in the velocity function, we can solve for t and that will be the exact time it takes the ball to reach its maximum height.

v = v_0y + at

0 = 42ft/sec - 32ft/sec² • t

-42ft/sec = -32ft/sec² • t

t = -42ft/sec / -32ft/sec²

t = 42/32 [ft•sec² / sec•ft]

t = 42/32 sec

Now we know the time when the ball reaches its maximum height, so we can plug it into our vertical position formula and obtain the height the ball reaches.

y = y₀ + v_0yt + 0.5at²

y = 0 + 42ft/sec• (42/32)sec + 0.5 • -32ft/sec² • (42/32)² sec²

y = 42²/32 ft + [(0.5 • -32 • 42²) / 32²] ft

y = 27.5625 ft

The maximum height the ball reaches is ~27.6 ft.