Question

A rock is thrown upward from the surface of Mars with an initial velocity of 50 feet per second. The height of a rock can be modeled by the: h(t)=-6.5t^2+50th(t)=−6.5t 2 +50t. How long does it take the rock to fall back to the surface of Mars? Round to two decimal places.

Answer 1

9.23secs

Explanation:

Given the height of the rock modeled by the equation;

h(t)=-6.5t^2+50t

The rock falls back to the surface of the Mars at when the height is zero i.e h(t) = 0

Substitute into the formula and get the time t

h(t)=-6.5t^2+50t

0 = -6.5t^2+50t

6.5t^2 = 50t

6.5t = 50

t = 50/6.5

t = 9.23secs

Hence the rock fall back to the surface of the Mars after 9.23secs