- #1

sarahcate

- 5

- 0

"A cricketer bowls a ball from a height of 2.3m. The ball leaves his hand horizontally with a velocity u. After bouncing once, it just passes 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 moment it is released the ball takes about 0.7s to fall 2.3m

b) How long does it take the ball to rise 0.71m after bouncing?"

Part a I'm ok with. I used v^2 = u^2 +2ax using

a = -9.81m/s2

x = 2.3m

u = 0m/s

to find v = -6.72 m/s

i then used v = u +at

to give t = (v-u)/a = 0.68s

For part b I thought I could use a similar approach saying that at the top of it's bounce the vertical component of the velocity will be zero and the only acceleration is due to gravity. So using v^2 = u^2 +2ax

with v= 0m/s

a = -9.81m/s2

x = 0.71

I got u = 3.73 m/s

and substituing that into t =(v-u)/a, t = 0.38s

unfortunately this was wrong and I got 0 out of a possible 3 marks.

Please can anyone help me with where I am going wrong? The only thing I can think of is that there is some additional acceleration in the vertical direction (from the bounce??) but I don't know how to work this out. If you can point me in the right direction that would be brilliant.