# Physics Trajectories QuestionWatch

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#1
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?
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1 year ago
#2
(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.
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1 year ago
#3
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
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1 year ago
#4
(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
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)
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1 year ago
#5
(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
I'm not berating this method, I'm just curious as well lol
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1 year ago
#6
(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)
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1 year ago
#7
(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.
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1 year ago
#8
(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
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#9
(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.
1
#10
(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
Thanks for the example! Helped me check my answer.
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