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Can someone help me with this question pleease (mechanics)

An object is projected vertically upwards, from a position 1.5m above horizontal ground, with speed 17.5m/s.
Calculate the height above ground level of the object when it is moving downwards at 15.1m/s. Give your answer to 3 significant figures.

I know it uses suvat equations (aka kinematic equations) but I keep getting the answer wrong, thanks in advance :biggrin:
Reply 1
Original post by ravioliii
An object is projected vertically upwards, from a position 1.5m above horizontal ground, with speed 17.5m/s.
Calculate the height above ground level of the object when it is moving downwards at 15.1m/s. Give your answer to 3 significant figures.
I know it uses suvat equations (aka kinematic equations) but I keep getting the answer wrong, thanks in advance :biggrin:

What did you try?
Reply 2
Original post by mqb2766
What did you try?

I already worked from a previous question that the maximum height above the ground would be 17.1m (15.6m above projection point).
So then I did suvat from this maximum height where s = ?, u = 0, v = 15.1, a = 9.8. (I also worked out the t=1.54)

v^2 = u^2 +2as -----> 15.1^2 = 0^2 + (2)(9.8)(s) and so s=11.6 to 3sf
This is 11.6m above the original projection point so add 1.5m to account for this = 13.1m above the ground
Reply 3
Original post by ravioliii
I already worked from a previous question that the maximum height above the ground would be 17.1m (15.6m above projection point).
So then I did suvat from this maximum height where s = ?, u = 0, v = 15.1, a = 9.8. (I also worked out the t=1.54)
v^2 = u^2 +2as -----> 15.1^2 = 0^2 + (2)(9.8)(s) and so s=11.6 to 3sf
This is 11.6m above the original projection point so add 1.5m to account for this = 13.1m above the ground

Id really sketch the problem to make sure youre using stuff consistently. The max height looks about right but then you do v=15.1 u=0 to get s=11.6 and this seems to be the displacement from the max.

You could just use u=17.5 and v=-15.1 to get the height above the launch part which is probably slightly simpler.
(edited 2 months ago)

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