Q: A rock is thrown off of a 100 foot cliff with an upward velocity of 45 m/s. As a result its height after t seconds is given by the formula:
h(t)=100+45t−4.9t2
(a)
What is its height after 3 seconds?
(b)What is its velocity after 3 seconds?
Answer:
(a) 190.9 m.
(b) 15.6 m/s upward
Explanation:
Given:
h(t) = 100 + 45t - 4.9t²
The height after 3 seconds,
t = 3 s
Substitute the value of t in to the equation above.
h(3) = 100+45(3)-4.9(3)²
h(3) = 100+135-44.1
h(3) = 190.9 m
Therefore the height after 3 seconds = 190.9 m.
(b) Velocity after 3 seconds
The velocity is obtained by differentiating h(t) with respect to time
v = dh(t)/dt
dh(t)/dt = 45-9.8t
v = 45 - 9.8t ......................................... Equation 1
t = 3 s.
Substitute the value of t into the equation above,
v = 45 - 9.8(3)
v = 45- 29.4
v = 15.6 m/s
Thus the velocity after 3 seconds = 15.6 m/s upward
