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A ball is thrown horizontally from a cliff and hits the ground 4 seconds later 40 meters from the base of the cliff how high was the cliff rounded to the nearest tenth of a meter?
Tauruschild1
Ignoring air resistance:
The distance from the base of the cliff is irrelevant as the only force acting on the ball is gravity acting straight down, giving it acceleration due to gravity.
s = ut + 1/2 at^2
For this problem: u = 0 m/s, t = 4 s, a = g (= acceleration due to gravity)
No value is given for g, but as an answer to the nearest tenth is required (1 dp), I'll use a value for g to 2 dp so that I can then round the answer to 1 dp: I'll use g ≈ 9.81 m/s^2
→ s ≈ 0 m + 1/2 x 9.81 m/s^2 x (4 s)^2 = 78.48 m
Rounded to the nearest tenth → 78.5 m.
horizontal distance = speed x time s = vt
45 = 15 t
t = 3 seconds
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