Posted by Pandy on Thursday, September 30, 2010 at 11:48pm.
What's given:
Cliff angle with the horizontal, θ = -19°
Distance on the cliff, d = 50m
Height of cliff, h = -50m
(0 being on top of the cliff)
acceleration before leaving cliff, a = 3.06 m/s²
acceleration due to gravity, g = -9.8 m/s²
Car was stationary at the beginning, so
initial velocity, u = 0
First calculate the speed of the car, v, when it left the cliff using
v²-u² = 2ad
At the point of leaving the cliff,
horizontal velocity, vh = vcos(θ)
vertical velocity, vv = vsin(θ)
(note the sign of θ)
We now can calculate the time in the air, t using
h = vv*t + (1/2)gt²
The horizontal distance from the cliff, S can be calculated by the product of the initial horizontal velocity, vh by the time in the air.
Post if you need explanations or answer check.
I'm not sure how to find a)
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