Question

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

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)