Answer:
The ball will go to maximum height of 56.25 meters.
Step-by-step explanation:
It is given that, a ball is thrown vertically upward from the ground. Initial velocity of the ball, v = 60 ft/sec
Initial height is 0 as the ball is thrown from ground. The equation of projectile is given by :
![h(t)=-16t^2+vt+h](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2Bvt%2Bh)
h = 0
..............(1)
We need to find the maximum height reached by the ball. For maximum height,
![\dfrac{dh(t)}{dt}=0](https://tex.z-dn.net/?f=%5Cdfrac%7Bdh%28t%29%7D%7Bdt%7D%3D0)
![\dfrac{d(-16t^2+60t)}{dt}=0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%28-16t%5E2%2B60t%29%7D%7Bdt%7D%3D0)
t = 1.875 seconds
Put the values of t in equation (1). So,
![h(t)=-16(1.875)^2+60(1.875)](https://tex.z-dn.net/?f=h%28t%29%3D-16%281.875%29%5E2%2B60%281.875%29)
h(t) = 56.25 meters
So, the ball will go to maximum height of 56.25 meters. Hence, this is teh required solution.