Question

An object is launched vertically in the air at 32.83 meters per second from a platform 10m above the ground. The height of the object above the ground (in meters) t seconds after launch is given by h(t)= -4.9t^2 + 32.83t +10 how long does it take the object to reach maximum height? What is the objects maximum height?

Answer 1

IdkAnswer:

Step-by-step explanation:

Answer 2

Answer: t = 3.35 s

h ≈ 65 meters

Step-by-step explanation:

The height h(t) give is the distance covered with respect to time t.

Recall : velocity = change in displacement/ change in time

And at the maximum height, it means that v = 0 , therefore dh / dt = 0

dh / dt = -9.8t + 32.83  ( differentiating h with respect to time)

equating it to zero , we have

- 9.8t + 32. 83 = 0

collecting the like terms

32.83 = 9.8t

Therefore

t = 32.83/9.8

t = 3.35 seconds

substitute the value of t in order to get the maximum height

h = -4.9($3.35^{2}$) + 32.83(3.35) + 10

= - 54.99025 + 109.9805+10

h = 64.99025

h≈ 65 meters