Problem 3.13: Leaping The River, I. A 10000 N car comes to a bridge during a sto

Question

Problem 3.13: Leaping The River, I.

A 10000 N car comes to a bridge during a storm and finds the bridge washed out. The 700 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 24.0 m above the river, while the opposite side is a mere 6.60 m above the river. The river itself is a raging torrent 63.0 m wide.

Part A: How fast should the car be travleing just as it leaves the cliff in order to clear the river and land safely on the other side?

Part B: What is the speed of the car just before it lands safely on the other side?

Explanation / Answer

The car must go from the higher side to the lower side, and make the 51 meters across before it falls farther than the height of the lower side. Acceleration due to gravity is 9.8m/s^2
Diff in heights, 24m - 6.6m = 17.4m = H
H = 1/2 a t^2 to find time to cross
17.4 = 1/2 (9.8) t^2 = 4.9 t^2
t^2 = 17.4/4.9 = 3.551,

t = 1.884 s, so horizontal
speed to cover 63m in 1.884s is d = vt, v = d/t = 63/1.884
v = 33.432 m/s
downward speed is v = at = 9.8 (1.884) = 18.463 m/s
total speed Vtot of car is sum of two vectors, which are added by Pythagorean theorem for right triangle:
(Vtot)^2 = (Vh)^2 + (Vd)^2 = (18.463)^2 + (33.432)^2
Vtot = sqrt (1458.588) = 38.19 m/s

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