A rock is thrown horizontally from a cliff with a speed of 15 m/s. It falls half the height of the cliff in the last three seconds of its fall. a. What is the total fall time?
b. How high is the cliff?
c. What is the horizontal distance from the cliff when it hits the ground?

A) 10.243 s

B) 514.64 m

C) 153.645 m

Explanation:

From projectile motion;

y(t) = h - ½gt²

Where;

h is the height of cliff

t is the time taken to fall from top of cliff down to half the height of the cliff

y(t) is height of the rock as function of time

We are told the rock falls half way of the cliff height.

Thus;

y = h/2

So;

h/2 = h - ½gt²

This gives;

h - h/2 = ½gt²

h/2 = ½gt²

h = gt² - - - - (1)

Also,the total free fall time from top of the cliff to ground would be;

T = t + 3

Thus, distance at this point is zero.

So;

0 = h - ½gT²

h = ½gT²

Putting t + 3 for T, we have;

h = (1/2)g(t + 3)² - - - 2

Inspecting eq 1 and eq 2,we have;

gt² = (1/2)g(t + 3)²

g will cancel out to give;

2t² = t² + 6t + 9

t² - 6t - 9 = 0

The roots are -1.243 or 7.243.

We will pick the positive value as time can't be negative.

Thus, T = 7.243 + 3 = 10.243 s

h = gt² = 9.81 × 7.243²

h = 514.64 m

Formula for horizontal distance is given by;

x = uT

Where u is initial speed given as;

u = 15 m/s

Thus;

x = 15 × 10.243

x = 153.645 m

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xd

50 inches should be the resultant vector