From the top of a cliff overlooking a lake, a person throws two stones. The two stones have identical initial speeds of v0 = 13.4 m/s and are thrown at an angle θ = 30.1°, one below the horizontal and one above the horizontal. What is the distance between the points where the stones strike the ground?

X = 15.88 m

Explanation:

Given:

Initial Velocity V₀ = 13.4 m/s

θ = 30.1 °

g = 9.8 m/s²

To Find horizontal distance let "X" we have to time t first.

so from motion 2nd equation at Height h = 0

h = V₀y t + 1/2 (-g) t ²                                (ay = -g)

0 = 13.4 sin 30.1° t - 0.5 x 9.81 x t²           (V₀y = V₀ Sin θ)

⇒  t = 1.37 s

Now For Horizontal distance  X, ax =0m/s²

X = V₀x t + 1/2 (ax) t ²

X = 13.4 m × cos 30.1° x 1.37 s + 0

X = 15.88 m

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