Question

A ball is thrown upward with an initial velocity of 60 mph. it is thrown from a height of 5 feet. what is the maximum height it reaches?

Answer 1

h = 61.25 m

Explanation:

It is given that,

The initial velocity of the ball, v = 60 m/s

It is thrown from a height of 5 feet,
`h_o=5\ ft`

We need to find the maximum height it reaches. The height reached by the projectile as a function of time t is given by :

`h=-16t^2+vt+h_0`

Putting all the values,

`h=-16t^2+60t+5` .....(1)

For maximum height, put

`(dh)/(dt)=0\\\\(-16t^2+60t+5)/(dt)=0\\\\-32t+60=0\\\\t=(-60)/(-32)\\\\t=1.875\ s`

Put t = 1.875 in equation (1)

`h=-16(1.875)^2+60(1.875)+5\\\\h=61.25\ m`

So, the maximum height reached by the ball is 61.25 m.