A rock is thrown horizontally off of a cliff. It starts with a velocity of 4.2 m/s and the cliff is 68.2 m high.
How far out from the bottom of the cliff does the rock fall?
57m
59m
63m
45.8m
To determine how far out from the bottom of the cliff the rock falls, we can use the kinematic equation for horizontal projectile motion:
𝑑 = 𝑣𝑥 × 𝑡
where:
𝑑 = distance traveled by the rock
𝑣𝑥 = horizontal velocity of the rock
𝑡 = time of flight
First, we need to find the time of flight. Since the rock is thrown horizontally, its initial vertical velocity is zero. We can use the kinematic equation for vertical motion under free fall to find the time it takes for the rock to reach the ground from a height of 68.2 m:
ℎ = 𝑣𝑦𝑜 × 𝑡 + (1/2) × 𝑔 × 𝑡²
where:
ℎ = initial vertical height (68.2 m)
𝑣𝑦𝑜 = initial vertical velocity (0 m/s)
𝑔 = acceleration due to gravity (-9.8 m/s²)
Simplifying the equation, we get:
68.2 = 0 × 𝑡 + (1/2) × (-9.8) × 𝑡²
68.2 = (-4.9) × 𝑡²
𝑡² = 68.2 / (-4.9)
𝑡² = -13.9184
𝑡 ≈ ±√(-13.9184)
Since time cannot be negative, we ignore the negative result for 𝑡.
𝑡 ≈ √(-13.9184) → This indicates that the rock does not fall back to the ground.
Therefore, the time of flight is not physically possible, and the rock does not fall out from the bottom of the cliff.
time to fall from height of clift:
68.2=1/2 g t^2
t= sqrt (68.2/4.9)=3.7 sec
distance horizontal= time*initial velocity