A ball is shot from the ground straight up into the air with initial velocity of

45 ft/sec. Assuming that the air resistance can be ignored, how high does it go? HINT: The acceleration due to gravity is 32 ft per second squared.

v = 45-32t

max height is when v=0
Now use that in
h = 45t - 16t^2

To determine the height the ball reaches, we can use the kinematic equation for vertical motion:

h = (v^2 - u^2) / (2g)

where:
h = height reached by the ball
v = final velocity of the ball (0 ft/sec at maximum height)
u = initial velocity of the ball (45 ft/sec)
g = acceleration due to gravity (32 ft/sec^2)

Plugging in the given values, we can calculate the height:

h = (0^2 - 45^2) / (2 * 32)
h = (0 - 2025) / 64
h = -2025 / 64
h ≈ -31.64 ft

Based on the calculations, we find that the ball reaches a height of approximately -31.64 feet. Keep in mind that this negative sign indicates that the height is measured above the ground level. Therefore, we can state that the height reached by the ball is approximately 31.64 feet.