Horizontal problem:
Speed = constant = U, the initial speed
distance = U t = 51.4 m
Vertical problem:
falls (16.6 - 7.9) = 8.7 meters DOWN
initial speed down = 0
so
8.7 = (1/2) (9.8) t^2
weight has no effect on this problem, only the acceleration of gravity
8.7 = 4.9 t^2
t^2 = 1.77
t = 1.33 seconds to9 fall
same t to reach the other bank so
U = 51.4 meters / 1.33 seconds
= 38.6 meters/sec
A 10,000N 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 16.6 m above the river, while the opposite side is a mere 7.90 m above the river. The river itself is a raging torrent 51.4 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?
This question has already been posted on here but I need to know how they found the initial velocity.
4 answers
Thank you so much.
You are welcome !
why is the initial speed down equal to 0?