A rock is thrown upward with a velocity

Question

A rock is thrown upward with a velocity of
25meters per second from the top of a
50meter high cliff, and it misses the cliff on the way back down. When will the rock be
7meters from the water, below? Round your answer to two decimal places.
 

Answer 1

h(t) = (1/2)at² + v₀t + h₀

 

Origin is at surface of water (h = 0), up is positive.

 

a = -g = -9.8 m/s²

v₀ = 25 m/s

h₀ = 50 m

 

h(t) = -4.9t² + 25t + 50

 

Find t when

 

7 = -4.9t² + 25t + 50

 

4.9t² - 25t - 43 = 0

 

t = [1/(2*4.9)] [25 ±√(25² + 4*4.9*43)]

 

t = (0.102)(25 ± 38.312)

 

t > 0 ⇒ use only the positive square root

 

t = (0.102)(25 + 38.312) ≅ 6.46 s

 

It is 7m above the water at about 6.46 s.

Answer 2

Use the equation

 

h = -4.9t² + 25t + 50

 

 

where h is the height above the water surface level.

 

 

Set  h=7 and solve for t.