h = 1/2g (v sinθ)^2
v sinθ t - 4.9t^2 = 0
r = v^2/g sin2θ
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.
4 answers
I don't understand any of this because I don't get how you would substitute the values into the equation?
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?
Is this correct?
How would I do the other two please?
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)
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)