Question

Two boys are throwing a baseball back and forth. The ball is 4 ft above the ground when it leaves one child’s hand with an upward velocity of 36 ft/s. If acceleration due to gravity is –16 ft/s^2, how high above the ground is the ball 2 s after it is thrown?

h(t) = at2 + vt + h0

Answer 1

`H(t) = at^(2) + vt + H_(0)`

`H(t) = -16t^(2) + 36t + 4`

`H(2) = -16(2)^2 + 36(2) + 4`

`H(2) = -64 + 72 + 4`

`H(2) = 12`

Therefore, two seconds after being thrown the ball is 12 feet above the ground.

Answer 2

Two seconds after being thrown the ball is 12 feet above the ground.

Explanation:

Given value,

Height = 4 ft

Time t = 2 sec

Velocity u = 36 ft/s

Using equation of motion.

`s=ut-(1)/(2)at^2+h_(0)`....(I)

Where, u = velocity

h = height

t = time

a = acceleration

Put the value in the equation (I)

`s=36*2-(1)/(2)*32*4+4`

`s =12\ m`

Hence, Two seconds after being thrown the ball is 12 feet above the ground.