Question

A rock is thrown off of a 110 foot cliff with an upward velocity of 50 ft/s. As a result its height after t seconds is given by the formula:h(t) = 110 + 50t - 5t^2What is its height after 2 seconds?___What is its velocity after 2 seconds? ____(Positive velocity means it is on the way up, negative velocity means it is on the way down.)

Answer 1

ANSWERS

• Height after 2 seconds: ,190 ft

,

• Velocity after 2 seconds: ,30 ft/s

EXPLANATION

The height of the rock is given by the equation,

$h(t)=110+50t-5t^2$

To find the rock's height after 2 seconds we have to replace t with 2 in the equation above and solve,

$h(2)=110+50\cdot2-5\cdot2^2=110+100-5\cdot4=210-20=190$

Hence, the height of the rock after 2 seconds is 190 ft.

The velocity of the rock is the derivative of the height,

$v(t)=h^{\prime}(t)=50-10t$

After 2 seconds,

$v(2)=50-10\cdot2=50-20=30ft/s$

Hence, the velocity of the rock after 2 seconds is 30 ft/s.