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 21.3 m above the river, whereas the opposite side is a mere 2.3 m above the river. The river itself is a raging torrent 54.0 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? 2. What is the speed of the car just before it lands safely on the other side?

Explanation:

1 ) Let the initial horizontal velocity of car be v .

For vertical displacement

vertical displacement h = 21.3 - 2.3 = 19 m

Time taken to fall by 19 m be t

19 = 1/2 x 9.8 t² ( initial downward velocity is zero )

t = 1.97 s

This is also the time taken to cover horizontal displacement of 54 m which is width of river .

horizontal speed v = 54 / 1.97 m /s

v = 27.41 m /s

2 )

At the time of landing on other side , car will have both vertical and horizontal speed .

vertical speed

v = u + gt

= 0 + 9.8 x 1.97 = 19.31 m /s

horizontal speed will remain same as  the initial speed = 27.41 m /s

Resultant speed = √ ( 27.41² + 19.31² )

= √ ( 751.3 + 372.87)

= 33.52 m /s

c. newton's third law of motion

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s