Question

An object is propelled off of a platform that is 75 feet high at a speed of 45 feet per second (ft/s). The height of the object off the ground is given by the formulah(0)= - 16t^2 + 45t + 75, where h (t) is the object's height at time (t) seconds after the object is propelled. The downward negative pull on the object is represented by -16t^2. Solve for t.

Answer 1

Answer:

t = 1.406 seconds

Explanation:

The given equation representing the height as a function of time t is:

$h(t)=-16t^2+45t+75$

Note that:

• The speed of the object at time t, v(t) = dh/dt

,

• At maximum height, v(t) = 0

dh/dt = -32t + 45

v(t) = -32t + 45

0 = -32t + 45

32t = 45

t = 45/32

t = 1.406 s