h = Hi + Vi t - 4.9 t^2
assume the road was level so Vi = 0
1.5 = 20.2 + 0 - 4.9 t^2
4.9 t^2 = 18.7
t = 1.95 seconds in the air to fall 18.7 meters
so how far did he go horizontal in those 1.95 seconds?
d = u t
60 = u * 1.95
u = 30.7 meters/second horizontal for the whole jump
what is v on the far side?
v = vi - g t
v = 0 - 9.8 (1.95)
v = - 19.1 m/s speed down (hard landing, car did not survive 18.7 meter fall of course)
speed at landing = sqrt (30.7^2+19.1^2)
= 36.2 m/s
(but it really did not land safely because the vertical landing speed was so high unless the road was sloped down at about 2 down for 3 horizontal on the far side)
A 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 20.2 m above the river, while the opposite side is a mere 1.5 m above the river. The river itself is a raging torrent 60.0 m wide.How fast should the car be traveling just as it leaves the cliff in order just 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?
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(I wonder if whoever gave you this problem really thought out the implications of the landing :)
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