car comes to a bridge during a storm and finds the bridge washed out. The driver

Question

car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 22.4 m above the river, whereas the opposite side is a mere 2.3 m above the river. The river itself is a raging torrent 53.0 m wide.

1.) How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?

Vo = ?  m/s

2.) What is the speed of the car just before it lands safely on the other side?

V = ?  m/s

I have no idea how to do this. Please show step by step with correct answers so I can try to understand.

THANK YOU!!

Explanation / Answer

d = at^2 / 2
(20.7 - 2.4) m = 9.8 m/s^2 * t^2 / 2
t^2 / 2 = 18.3m / 9.8 m/s^2
t^2 = 2(18.3 / 9.8) = 1.87 s
t = 1.37 s

d = vt
v = d / t; where t = 1.37 s
v = 66 m / 1.37 s
v = 48.2 m/s or 173.5 km/h is the required speed of the car.

Velocity of car when it lands is the combined horizontal velocity plus the velocity of its accelerated drop:

v(y) = at
v(y) = 9.8 m/s^2 * 1.37s = 134.26 m/s

v(combined) = (48.2 + 134.26) m/s = 182.5 m/s

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