A cricket ball is thrown at a speed of 30ms^-1 in a direction 30 degrees above the horizontal. Calculate:

Question

1. The maximum height.

2. The time taken to return to the same level.

3. The distance from the thrower to where the ball returns to the same level.

Answer 1

h = 1/2g (v sinθ)^2

v sinθ t - 4.9t^2 = 0

r = v^2/g sin2θ

Answer 2

I don't understand any of this because I don't get how you would substitute the values into the equation?

Answer 3

I would assume g as being 10ms^-2 therefore for the maximum height, I got: h = 1/2g (v sinθ)^2 =0.5×10^-2(30sin(30))^2 =1.125

Is this correct?

How would I do the other two please?

Answer 4

Here,velocity of projection,u=30 m/s
Angle of projection=30°
We have,maximum height attained by a projectile is
H=u^2 sin tita/2g (u square sin tita divided by 2g)
=30^2×sin 30/2×9.8
=900×1/2/(whole divided by)19.6
=450/19.6
=22.95
=23 m (approx)