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

Step-by-step explanation:

This question is part of a graded assignment given by a cyber school in Pennsylvania. I know this to be true because I can see the next question as well in the screen shot. Please remove this question and answer. Signed, the teacher.

P.S. The phase (when height is zero) was added to the original question provided by the curriculum provider so students solve for the time in the air, and not the time required to reach maximum height.

Answer 2

Answer:

t = 1.406 seconds

Step-by-step 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