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?

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() + 32.83(3.35) + 10

= - 54.99025 + 109.9805+10

h = 64.99025

h≈ 65 meters

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq

where did you get them? if you picked from the grass, is is probably poisonous.

step-by-step explanation: