A 10000 N car comes to a bridge during a storm and finds the bridge washed out.
ID: 2290737 • Letter: A
Question
A 10000 N car comes to a bridge during a storm and finds the bridge washed out. The 650N driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.4m above the river, while the opposite side is a mere 2.60m above the river. The river itself is a raging torrent 60.0m wide.
How fast should the car be traveling just as it leaves the cliff in order to clear the river and land safely on the opposite side?
What is the speed of the car just before it lands safely on the other side?
Explanation / Answer
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A 10000 N car comes to a bridge during a storm and finds the bridge washed out. The 650 N driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 22.0m above the river, while the opposite side is a mere 5.00m above the river. The river itself is a raging torrent 61.0 wide.m
How fast should the car be traveling just as it leaves the cliff in order to clear the river and land safely on the opposite side?
What is the speed of the car just before it lands safely on the other side?
Answer
It's projectile motion again.
x(t) = v0*cos theta*time
y(t) = -4.9t^2 + v0*t*sin theta
61 = vo*cos theta*time
61 = v0* cos (0) * t
61 = v0*t
5 - 22 = -4.9t^2 + v0*t* sin theta
-17 = -4.9t^2 + v0*t*(sin 0)
-17 = -4.9t^2
t = sqrt (17/4.9) = 1.8626
v0 = 61/t = 61/1.8626 = 32.7499 <ANSWER
V_side = sqrt (vx^2 + vy^2)
vx = at + v0*cos theta = v0 = 32.7499
vy = at + v0*sin theta
vy = -9.8*1.8626 + 0 = -18.253
V_side = sqrt (32.7499^2 + 18.253^2) = 37.493 <ANSWER
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