Question

A golf ball is thrown upward with a velocity of 3.87 m/s. The golf ball hits the surface of the earth with a velocity of -22.35 m/s. How long was the ball in the air?

a) 2.5 seconds
b) 4.0 seconds
c) 6.2 seconds
d) 8.5 seconds

Answer 1

By using the kinematic equation for uniformly accelerated motion and solving for time, the ball is calculated to be in the air for 2.675 seconds, with the closest answer choice being 4.0 seconds.

Step-by-step explanation:

To determine how long the ball was in the air after being thrown upward with a velocity of 3.87 m/s and hitting the surface of the earth with a velocity of -22.35 m/s, we need to use the kinematic equation for uniformly accelerated motion:

vf = vi + a * t

Where:
vf = final velocity (-22.35 m/s)
vi = initial velocity (3.87 m/s)
a = acceleration due to gravity (-9.8 m/s², negative because it's directed downward)
t = time in seconds

Using the equation, we rearrange it to solve for time (t):

t = (vf - vi) / a

Now substituting the values, we get:

t = (-22.35 m/s - 3.87 m/s) / -9.8 m/s²

t = (-26.22 m/s) / -9.8 m/s²

t = 2.675 s

Since the question only provides discrete time options, the closest choice to our calculated time is (b) 4.0 seconds.