A car initially at rest slides down a steep hillside and then falls off a cliff.

Question

A car initially at rest slides down a steep hillside and then falls off a cliff. On the hillside (which can be treated as one-dimensional motion since it is a straight line), assume that the car has an acceleration whose magnitude is 4 m/s2 . Once it leaves the cliff, the car is certainly undergoing projectile motion and is subject only to gravity. (a) Determine the distance it lands from the cliff R. Use D = 50 m, Theta = 30 degrees, and H = 100 m. (b) Graph horizontal position, vertical position, horizontal velocity, vertical velocity, horizontal acceleration and vertical acceleration as a function of time for the time period when it is in projectile motion.

Explanation / Answer

I will assume D is the length of the hill the car slid, and H is the height of the hill, if these are not correct assumptions, please make the numerical changes as needed in this procedure

first, find the speed of the car when it reaches the edge of the cliff

use vf^2 = v0^2 + 2 ad

vf = final speed, v0 = initial speed = 0, a = 4 m/s2, d = 50 m

vf^2 = 0 + 2 x 4 m/s2 x 50 m
vf = 20 m/s

the car leaves the edge of the cliff with an initial velocity of 20m/s; the horizontal component of this is

20 cos 30 = 17.32 m/s

the vertical component = - 20 sin 30 = -10 m/s where the negative sign indicates motion downward

we find the time of flight from

y(t) = y0 + v0y t - 1/2 g t^2

y0=initial height = 100m; v0y = initial vertical velocity = - 10 m/s, g = 9.8 m/s2

we find the time of flight by setting y(t) =0 and solving the quadratic

0 = 100 - 10t - 4.9t^2

t=3.61s

finally, we find the range from

R = horizontal velocity x time of flight = 17.32 m/s x 3.61 s = 62.53 m

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