Respuesta :
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 :
[tex]h(t)=-16t^2+vt+h[/tex]
h = 0
[tex]h(t)=-16t^2+60t[/tex]..............(1)
We need to find the maximum height reached by the ball. For maximum height,
[tex]\dfrac{dh(t)}{dt}=0[/tex]
[tex]\dfrac{d(-16t^2+60t)}{dt}=0[/tex]
t = 1.875 seconds
Put the values of t in equation (1). So,
[tex]h(t)=-16(1.875)^2+60(1.875)[/tex]
h(t) = 56.25 meters
So, the ball will go to maximum height of 56.25 meters. Hence, this is teh required solution.
