A rock is thrown horizontally from a cliff with a speed of 15 m/s. It falls 1/2 of the cliff height in the last three seconds of its fall. Find free fall time.

Question

Also find the find the cliff height and the horizontal distance from the cliff. 

Answer 1

Chris,

 

I will set this up for you, but you will have to grind through the math.

 

h = height of cliff

T = total free-fall time from the top of the cliff to ground level

t₁ = time to fall from top of cliff down to half the height of the cliff

 

y(t) = height of rock as function of time

 

y(t) = h - (1/2)gt²

 

g = 9.8 m/s²

 

h/2 = h - (1/2)gt₁²

 

T = total free fall time from y = h to y = 0

 

h = (1/2)gT²

 

T = t₁ + 3  [seconds]

 

You have 3 unknowns h, T, and t_1; and 3 independent equations to work with: 

 

h = (1/2)gT²

h/2 = (1/2)gt₁²

T = t₁ + 3

 

Solve for h, T, and t₁.  (The algebra will be tedious.  Good luck if you have to solve by hand!)

 

Once you have T, get the horizontal travel distance from

 

x = v_0xT

 

using your solution to T from above, and

 

v_0x = 15 m/s

Answer 2

From the information given in the problem (1/2)(y) = 3 Or y=6 y=(1/2)gt^2 (6/4.9)=t^2 t=1.106 sec x=vt x=(15)(1.106)=16.59 m