# Projectiles q

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#1
Hi can someone check my working out for this question please?ty
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1 month ago
#2
a) It goes 2km into the air? You may want to rethink "v" (and u). Also 0.5 on the denominator?
b) is a distance, so having "s" units is courageous. The value for t, which is part of the solution, isnt correct. Linking part a and b), you'd get to a height of 2km in ~1.3s.
Last edited by mqb2766; 1 month ago
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#3
(Original post by mqb2766)
a) It goes 2km into the air? You may want to rethink "v" (and u). Also 0.5 on the denominator?
b) is a distance, so having "s" units is courageous. The value for t, which is part of the solution, isnt correct. Linking part a and b), you'd get to a height of 2km in ~1.3s.
Hi Sorry for the late response, I did part a again after realising some really silly mistakes and this is what I got
Last edited by akawesome; 1 month ago
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1 month ago
#4
(Original post by akawesome)
Hi Sorry for the late response, I did part a again after realising some really silly mistakes and this is what I got
For part a, the initial vertical velocity, u, will be 100 sin 25, not cos. Although, you've actually got the correct answer, so a typo?
Last edited by ghostwalker; 1 month ago
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1 month ago
#5
(Original post by akawesome)
Hi Sorry for the late response, I did part a again after realising some really silly mistakes and this is what I got
Part b). Same comments as before. The answer is (should be) a distance and the time calculation is wrong. 2 s to go 100m up and down is too short. You can check by subbing into the previous formula and 2*5 (roughly) is not 100sin(25).
Last edited by mqb2766; 1 month ago
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#6
(Original post by ghostwalker)
For part a, the initial vertical velocity, u, will be 100 sin 25, not cos. Although, you've actually got the correct answer, so a typo?
Yeah sorry
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#7
(Original post by mqb2766)
Part b). Same comments as before. The answer is (should be) a distance and the time calculation is wrong. 2 s to go 100m up and down is too short. You can check by subbing into the previous formula and 2*5 (roughly) is not 100sin(25).
Its travelling at 100m/s so wouldnt t be in seconds anyway?
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1 month ago
#8
(Original post by akawesome)
Its travelling at 100m/s so wouldnt t be in seconds anyway?
You seem to be assuming constant vertical speed this time?
There are a few ways to do it. Why not write down clearly what your assumptions are and which suvat etc.
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#9
(Original post by mqb2766)
You seem to be assuming constant vertical speed this time?
There are a few ways to do it. Why not write down clearly what your assumptions are and which suvat etc.
Im really sorry i keep bothering you but i did it again and i wrote all my assumptions and i think my reasoning should be clearer
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1 month ago
#10
Minor quibbles about the sig figs and the t=0 is the solution when the motion starts. But that looks better.

How to check/do it differently? Initial vertical velocity is ~42, v=u+at, so time to peak = 42/9.8 ~ 4.3s. Double to get range time.

You used the vertical displacement quadratic, which is also correct, but more complex. The back of an envelope calculation is that it takes 4s for a velocity of 40 to go to 0 decelerating at 10m/s^2. Simple approximations/validations stops gross errors.
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#11
(Original post by mqb2766)
Minor quibbles about the sig figs and the t=0 is the solution when the motion starts. But that looks better.

How to check/do it differently? Initial vertical velocity is ~42, v=u+at, so time to peak = 42/9.8 ~ 4.3s. Double to get range time.

You used the vertical displacement quadratic, which is also correct, but more complex. The back of an envelope calculation is that it takes 4s for a velocity of 40 to go to 0 decelerating at 10m/s^2. Simple approximations/validations stops gross errors.
Thank you! I actually ordered a book called maths onthe back of an envelope to help me estimation so hopefully I'll get better but thanks alot!
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1 month ago
#12
(Original post by akawesome)
Thank you! I actually ordered a book called maths onthe back of an envelope to help me estimation so hopefully I'll get better but thanks alot!
Also, if you're using suvat varibles for both horizontal and vertical motion, it may help to use s_x, s_y, u_x, u_y, ... so you dont get confused. You confused acceleration being zero vertically in the second attempt so a_x = 0, a_y = -9.8.
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#13
(Original post by mqb2766)
Also, if you're using suvat varibles for both horizontal and vertical motion, it may help to use s_x, s_y, u_x, u_y, ... so you dont get confused. You confused acceleration being zero vertically in the second attempt so a_x = 0, a_y = -9.8.
😀👍
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