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Physics Trajectories Question

Stuck on this trajectory question on Isaac Physics, any help would be appreciated thanks:
A cricket batsman hits a ball at a speed of 27m/s at 60∘ to the horizontal. How far away would you have to stand in order to catch it?
Original post by Hughcifer
Stuck on this trajectory question on Isaac Physics, any help would be appreciated thanks:
A cricket batsman hits a ball at a speed of 27m/s at 60∘ to the horizontal. How far away would you have to stand in order to catch it?


Steps:

a) Resolve the velocity vector into vertical and horizontal components.

b) Then use a suitable SUVAT to work out how long it takes the ball to reach the apex of it's trajectory using the vertical component.

c) What goes up takes the same duration to come down. So use the time calculated for the total flight together with the horizontal speed component to calculate how far the ball travels.
(edited 6 years ago)
Reply 2
Hi, sorry my handwriting is a mess, I’ve done the question below (I hope it’s correct), if you have any questions, feel free to ask me :smile:
D3296C2F-9713-4966-AD1E-A32049A331C4.jpg
Original post by lewisli_
Hi, sorry my handwriting is a mess, I’ve done the question below (I hope it’s correct), if you have any questions, feel free to ask me :smile:
D3296C2F-9713-4966-AD1E-A32049A331C4.jpg


Wouldn't it be easier to just use v^2 = u^2 + 2as and solve for s? (multiply through by 2 because it's at maximum height)
Original post by lewisli_
Hi, sorry my handwriting is a mess, I’ve done the question below (I hope it’s correct), if you have any questions, feel free to ask me :smile:
D3296C2F-9713-4966-AD1E-A32049A331C4.jpg


I'm not berating this method, I'm just curious as well lol
Reply 5
Original post by AryanGh
Wouldn't it be easier to just use v^2 = u^2 + 2as and solve for s? (multiply through by 2 because it's at maximum height)


I’m not sure what you mean by multiplying by 2 but I don’t think it would work for solving for s because we‘re using the vertical component. The acceleration due to gravity is also in the vertical component and the ball travelled in the horizontal component too?? (If that makes sense)
Original post by lewisli_
I’m not sure what you mean by multiplying by 2 but I don’t think it would work for solving for s because we‘re using the vertical component. The acceleration due to gravity is also in the vertical component and the ball travelled in the horizontal component too?? (If that makes sense)


Ah. I see what you mean kinda. I was guessing that because the ball, a projectile, is assumed to have parabolic motion in the question. I thought you could just find it's maximum height which gives half the displacement, then multiply by 2 for the whole thing.
Reply 7
Original post by AryanGh
Ah. I see what you mean kinda. I was guessing that because the ball, a projectile, is assumed to have parabolic motion in the question. I thought you could just find it's maximum height which gives half the displacement, then multiply by 2 for the whole thing.


oh i see, ye i’m not good at explaining loool but it’s because of the horizontal component. If the ball was thrown vertically upwards and you wanted to find the distance travelled by the ball altogether, I guess what you’re saying would work. though from the components that i resolved, the vertical and horizontal initial velocities are different
Reply 8
Original post by uberteknik
Steps:

a) Resolve the velocity vector into vertical and horizontal components.

b) Then use a suitable SUVAT to work out how long it takes the ball to reach the apex of it's trajectory using the vertical component.

c) What goes up takes the same duration to come down. So use the time calculated for the total flight together with the horizontal speed component to calculate how far the ball travels.


Thanks for that one! Should be alright with these questions now.
Reply 9
Original post by lewisli_
Hi, sorry my handwriting is a mess, I’ve done the question below (I hope it’s correct), if you have any questions, feel free to ask me :smile:
D3296C2F-9713-4966-AD1E-A32049A331C4.jpg


Thanks for the example! Helped me check my answer.

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