During a storm, a car travelling on a level horizontal road comes upon a bridges

Question

During a storm, a car travelling on a level horizontal road comes upon a bridges that has washed out. The driver must get to the other sides, so he decides to try leaping the river with his car. The side of the road the car is on is 22.8 m above the river, while the opposite side is only 1.9 m above the river. The river itself is a raging torrent 55.0 in wide. How fast should the car be travelling at the time it leaves the road in order just to clear the river and land safely on the opposite side? Express your answer with the appropriate units. What is the speed of the car just before it lands on the other side? Express your answer with appropriate units.

Explanation / Answer

A)
Let the speed with which it was traveeling be v m/s

for vertical direction:
vi = 0
d = -(22.8-1.9)m = -20.9 m
a = -9.8 m/s^2

use:
d = vi*t + 0.5*a*t^2
-20.9 = 0 + 0.5*(-9.8)*t^2
t = 2.07 s

Vf = Vi + a*t
= 0 - 9.8*2.07
= -20.24 m/s

for horizontal direction:
v is constant
t = 2.07 s
d = 55 m
v = d/t
= 55/2.07
= 26.63 m/s

Answer: 26.63 m/s

B)
final velocity on horizontal direction will not chnage
It will be 26.63 m/s

In vertical direction, final velocity = -20.24 m/s < ---As calculated above

speed= sqrt (26.63^2 + 20.24^2)
= 33.45 m/s
Answer: 33.45 m/s

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