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since the ball's initial velocity after collision is perfectly horizontal, take the horizontal distance moved and divide it by the time above(1.333)
horizontal velocty stays constant throughout travel.
3.b.i v^2=u^2 - 2gs
--> 0= 13^2 - (2*9.81 s)
--> s= 8.61 ~ 8.61m
3.b.ii v= u -gt Alternatively, you can also use horizontal velocity to calculate the time.
--> 0= 13 - 9.81*t time = distance/ horizontal velocity = 9.95/7.5 = 1.33 s
--> t= 1.325 s ~ 1.33 s
3. b.iii using vertical velocity, The horizontal velocity IS THE SPEED after rebounding from P. There is no vertical speed at P since in the beginning of the question
--> s= ut + .5 gt^2 it was said "The ball hits the wall at P with a velocity that is at right angles to the wall" This means there is no vertical velocity at P.
--> 8.61 = 4.905 t^2
--> t = 1.324 s
so horizontal velocity = 6.15/1.324 = 4.64 ~ 4.6 m/s
Hope that helps man.
ohh ok
thanks