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

Answer:

h = 61.25 m

Step-by-step 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

$\dfrac{dh}{dt}=0\\\\\dfrac{-16t^2+60t+5}{dt}=0\\\\-32t+60=0\\\\t=\dfrac{-60}{-32}\\\\t=1.875\ s$

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.