Answer:
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
