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Mechanics - SUVAT help

4. A particle P1 is projected vertically upwards,from a horizontal ground with a speed of 30ms^-1. At the same instant another particle P2 is projected vertically upwards from the top of a tower of height 25m with a speed of 10ms^-1.

c) the time for which P1 is higher than P2 and is moving upwards.

How would you go about solving this question?
Original post by SmartFailure
4. A particle P1 is projected vertically upwards,from a horizontal ground with a speed of 30ms^-1. At the same instant another particle P2 is projected vertically upwards from the top of a tower of height 25m with a speed of 10ms^-1.

c) the time for which P1 is higher than P2 and is moving upwards.

How would you go about solving this question?


Its hard to explain with words through text on a computer so I'll just tell you it will really help if you drew a Velocity-Time Graph for P1 and P2. Also Important, Remember the distance between P1 and P2 is 25m as the top of the tower has a height of 25m.
(edited 6 years ago)
Original post by SmartFailure
4. A particle P1 is projected vertically upwards,from a horizontal ground with a speed of 30ms^-1. At the same instant another particle P2 is projected vertically upwards from the top of a tower of height 25m with a speed of 10ms^-1.

c) the time for which P1 is higher than P2 and is moving upwards.

How would you go about solving this question?


It's been a little while since I've done this but I think you would use the equation s-s0=ut + 1/2 at^2 for both? It is easier if you use the point from which P1 is projected as the origin so that s0 = 0 in the equation for P1, and 25 for P2. Then compare the two equations.
A particle, P1, is projected vertically upwards, from horizontal ground, with a speed of
30ms-1. At the same instant another particle, P2, is projected vertically upwards from the top of a tower of height 25m, with a speed of 10ms-1.

Find the time for which P1 is higher than the top of the tower.

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