A catapult launches a rock with an initial speed of 35 m/s at an initial angle of 22o above the ground from a height of 0.0 m. After the rock has traveled 10 m horizontally along the ground from its starting point, what is its height above the ground

h = 3.6 m

Explanation:

This is the case of a projectile motion. We assume the air resistance to be negligible. So the speed of the rock remains constant in x-direction. Therefore,

s = V₀ₓ t

where,

V₀ₓ = x - component of launching velocity = V Cos θ = (35 m/s)(Cos 22°)    V₀ₓ = 32.45 m/s

s = distance traveled horizontally by the rock = 10 m

t = time taken to cover the distance = ?

Therefore,

10 m = (32.45 m/s)(t)

t = (10 m)/(32.45 m/s)

t = 0.31 s

Now, we consider the vertical motion. The vertical motion is taking place under the action of gravity. So it is uniformly accelerated motion. Applying 2nd equation of motion to the vertical motion:

h = (Vi₀)(t) + (0.5)gt²

where,

h = height of rock = ?

Vi₀ = Vertical Component of Launching Velocity = V₀Sinθ = (35 m/s)(Sin 22°)

Vi₀ = 13.11 m/s

g = - 9.8 m/s² (negative for upward motion)

Therefore,

h = (13.11 m/s)(0.31 s) + (0.5)(-9.8 m/s²)(0.31 s)²

h = 4.06 m - 0.47 m

h = 3.6 m

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