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# Maths M1 edexcel question Watch

1. 2. A small stone is projected vertically upwards from a point O with a speed of 19.6msí1.
Modelling the stone as a particle moving freely under gravity,
(a) find the greatest height above O reached by the stone,
(2)
(b) find the length of time for which the stone is more than 14.7 m above O.

I don't understand why v= 19.6 for part b, shouldnt it be 0 at the maximum height.

Thanks in advance
2. (Original post by James_pa)
2. A small stone is projected vertically upwards from a point O with a speed of 19.6msí1.
Modelling the stone as a particle moving freely under gravity,
(a) find the greatest height above O reached by the stone,
(2)
(b) find the length of time for which the stone is more than 14.7 m above O.

I don't understand why v= 19.6 for part b, shouldnt it be 0 at the maximum height.

Thanks in advance

I think you're confused. This has nothing to do with the maximum height. Imagine the path of the projectile, it goes up, crosses the 14.7 metre mark (start your stopwatch!) continued upwards, hits the maximum height, start moving downwards, crosses the 14.7 metre mark again (stop your stopwatch!) and then hits the ground.

You want the time that your stopwatch has recorded. Nothing to do with the maximum height, picture below:

Standard way to do this is to find the times at which the particle is at 14.7 and then subtract the two times.

To do so, set 14.7 = ut + 1/2at^2 and then solve the quadratic equation in t to get two solutions. Bigger solution - smaller solution = time.

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