Asked by Tom Lisby
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.
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.
Answers
Answered by
Steve
h = 1/2g (v sinθ)^2
v sinθ t - 4.9t^2 = 0
r = v^2/g sin2θ
v sinθ t - 4.9t^2 = 0
r = v^2/g sin2θ
Answered by
Saphina AlMatary
I don't understand any of this because I don't get how you would substitute the values into the equation?
Answered by
Saphina AlMatary
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?
Answered by
Sagarika
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)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.