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

Question

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 19.4 m above the river, whereas the opposite side is a mere 2.2 m above the river. The river itself is a raging torrent 69.0 m wide.
Part A. 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?
Part B. What is the speed of the car just before it safely lands on the other 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 19.4 m above the river, whereas the opposite side is a mere 2.2 m above the river. The river itself is a raging torrent 69.0 m wide.
Part A. 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?
Part B. What is the speed of the car just before it safely lands on the other 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 19.4 m above the river, whereas the opposite side is a mere 2.2 m above the river. The river itself is a raging torrent 69.0 m wide.
Part A. 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?
Part B. What is the speed of the car just before it safely lands on the other side?

Explanation / Answer

Initial Horizontal speed = u
Initial Vertical speed = 0

Vertical Distance car needs to travel = Difference in Height of two bridge sides = 19.4 - 2.2 = 17.2 m
Horizontal Distance car needs to travel = Width of River = 69.0 m

Horizontal Acceleration = 0
Vertical Acceleration =9.8 m/s^2 downwards

Time taken to carry Horizontal Distance = Time taken to carry Vertical Distance

Time taken to carry Horizontal Distance =
S = u*t + 0.5 *at^2
69 = u *t ----------------1

Time taken to carry Vertical Distance =
s = u *t + 0.5 * a *t^2
17.2 = 0 + 0.5 *9.8*t^2
t = sqrt(17.2/4.9)
t = 1.87 s

Part A -
Speed of car before leaving cliff u = 69/1.87 m/s
Speed of car before leaving cliff u = 36.9 m/s

Part B -
Horizontal speed u = 36.9 m/s
Vertical speed v = 9.8 * 1.87 = 18.33 m/s
Speed of car on landing = sqrt(36.9^2 + 18.33^2 ) m/s
Speed of car on landing = 41.2 m/s

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