I was practicing exams for PHY 1 but I don't understand how to do this question, even with the mark scheme.
A cricketer bowls a ball from a height of 2.3m. The ball leaves the hand horizontally with a velocity u. After bouncing once, it passes just over the stumps at the top of its bounce. The stumps are 0.71m high and are situated 20m from where the bowler releases the ball.
a) Show that from the momnt it is released, the ball takes about 0.7 s to fall 2.3 m.
This is pretty easy, resolving vertically,
u = 0 m/s
a= -9.81 m/s/s
s= 2.3 m
Using v^2= u^2 + 2as
u^2 cancels out.
2 x -9.81 x 2.3 = 45.126
Square root ans = v = -6.72 m/s
Using t = v-u/a
-6.72-0/-9.81 = 0.685s
Then it says, b) How long does it take the ball to rise 0.71 m after bouncing? (3 mark question)
Then c) Use your answers to parts (a) and (b) to calculate the initial horizontal velocity u of the ball. You may assume that the horizonatl velocity has remained constant (2 marks)
d) In reality the horixontal velocity would not be constant. State one reason why.
The answer to this is simple too: There will be significant air resistance at the bounce.
Please could someone help with parts b and c.