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?

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