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**djmccabie** A ball is kicked from a point A on the horizontal ground with an initial speed of 24.5 m/s at an angle a above the horizontal, where sina = 0.6. The ball hits the ground at point B.

a) Calculate the time of flight of the ball

b) Find the horizontal distance AB

c) Find the speed of the ball when it is at a height of 9.6 above the ground.

OK for a this is what I do but not sure if its right.

Vy = vsina - 9.8t

Vy = 24.5*0.6-0

Vy = 14.7

r = vsin(a)t - 4.9t²=0

r = 14.7*0.6t - 4.9t²=0

using the formula

(8.82+8.82)/9.8 = 1.8s

Could anybody confirm if this is done correctly?

for part b i get

AB = vsin(a)t - 4.9t²

=24.5*0.6*1.8-4.9(1.8)²

=26.46 - 15.876

= 10.584m

Don't know how to start part C :/