# Projectiles q

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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.

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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(Original post by

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.

**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.

Last edited by akawesome; 1 month ago

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#4

(Original post by

Hi Sorry for the late response, I did part a again after realising some really silly mistakes and this is what I got

**akawesome**)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 ghostwalker; 1 month ago

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#5

**akawesome**)

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 mqb2766; 1 month ago

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(Original post by

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?

**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?

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(Original post by

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).

**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).

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#8

(Original post by

Its travelling at 100m/s so wouldnt t be in seconds anyway?

**akawesome**)Its travelling at 100m/s so wouldnt t be in seconds anyway?

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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(Original post by

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.

**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.

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#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.

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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(Original post by

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.

**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.

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#12

(Original post by

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!

**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!

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(Original post by

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.

**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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