A stone is thrown horizontally with a speed of 15 m/s from the top of a vertical cliff at the edge of a lake. If the stone hits the water 2.0 s later, the height of the cliff is closest to:

a. 90.0 m
b. 80.0 m
c. 100 m
d. 110 m

h= 43.07 m

Explanation:

Given:

Vi = 15 m/s, Vf= 0m/s, t = 2s, a=g=9.8 m/s²

To Find:

height of the clip h= ?m

Solution:

Since it is thrown horizontally  an angle of 0°, so Viy = Viy sin 0° =  0 m/s

and Vix = Vix cos 0° =  15 m/s

For Horizontal

h = 0 m, ax = 0m/s and Vix = 15 m/s

so h = h₀ + Vix t + 1/2 a t²

⇒ h₀ = 30 m  (Horizontal distance)

For vertical

Viy = 0m/s, h= ? m, a =g=9.8 m/s²

h = h₀ + Viy t + 1/2 a t²

h = 30m + 0 + 1/2 (9.8 m/s² × 2²)

h = 30 + 16.07

h= 43.07 m

seeing the graph would

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