Hughcifer
Badges: 9
Rep:
?
#1
Report Thread starter 1 year ago
#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?
0
reply
uberteknik
  • Study Helper
Badges: 21
Rep:
?
#2
Report 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.
0
reply
lewisli_
Badges: 7
Rep:
?
#3
Report 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
Name:  D3296C2F-9713-4966-AD1E-A32049A331C4.jpg
Views: 185
Size:  287.3 KB
0
reply
thotproduct
Badges: 19
Rep:
?
#4
Report 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
Name:  D3296C2F-9713-4966-AD1E-A32049A331C4.jpg
Views: 185
Size:  287.3 KB
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)
0
reply
thotproduct
Badges: 19
Rep:
?
#5
Report 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
Name:  D3296C2F-9713-4966-AD1E-A32049A331C4.jpg
Views: 185
Size:  287.3 KB
I'm not berating this method, I'm just curious as well lol
0
reply
lewisli_
Badges: 7
Rep:
?
#6
Report 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)
0
reply
thotproduct
Badges: 19
Rep:
?
#7
Report 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.
0
reply
lewisli_
Badges: 7
Rep:
?
#8
Report 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
0
reply
Hughcifer
Badges: 9
Rep:
?
#9
Report Thread starter 1 year ago
#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
reply
Hughcifer
Badges: 9
Rep:
?
#10
Report Thread starter 1 year ago
#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
Name:  D3296C2F-9713-4966-AD1E-A32049A331C4.jpg
Views: 185
Size:  287.3 KB
Thanks for the example! Helped me check my answer.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Bradford
    Undergraduate and Postgraduate Open day Postgraduate
    Thu, 24 Oct '19
  • Cardiff University
    Undergraduate Open Day Undergraduate
    Sat, 26 Oct '19
  • Brunel University London
    Undergraduate Open Day Undergraduate
    Sat, 26 Oct '19

Would you turn to a teacher if you were being bullied?

Yes (69)
23.71%
No (222)
76.29%

Watched Threads

View All