Question

A ball is kicked upward with an initial velocity of 68 feet per second. The ball's height, h (in feet), from the ground where it is measured in seconds. How much time does the ball take to reach its highest point? What is its height at this point?

Answer 1

Acceleration due to gravity is 32.2 ft/s^2

68=32.2t
t=2.11 seconds

h=68t-32.2/2*t^2
h(2.11)=68*2.11-32.2/2*2.11^2
h=71.8 feet

Answer 2

Answer:

72.25 ft.

Step-by-step explanation:

When the ball would reach the maximum height, its velocity would be zero. Acceleration due to gravity would act downwards = -32 ft/s²

Using the first equation of motion, time can be calculated as:

v - u = a t

⇒ 0-68 ft/s = - 32 ft/s²× t

⇒t = 2.125 s

The ball would take 2.125 s to its highest point.

Using the third equation of motion, height can be calculated as:

$s=\frac{v^2-u^2}{2a}\\ \Rightarrow h = \frac{0-(68ft/s)^2}{2\times 32ft/s^2} = 72.25 ft$

Thus, its height would be 72.25 ft.