Would appreciate if someone could help on this suvat question, part 3.
A particle P is projected vertically downwards from a fixed point O with initial speed 4.2ms−1, and takes 1.5 s to reach the ground. Calculate
(i) the speed of P when it reaches the ground,
(ii) the height of O above the ground,
(iii) the speed of P when it is 5m above the ground.
I'm stuck on the third part of this question.
I got S as 5 , u as 4.2 , v as v , a as 9.8 .
V^2=U^2+2as , v= root(4.2^2+98) , v=10.7 which is wrong
Thanks , saw where the error was
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Last edited by Tulian; 28-05-2012 at 21:43.
- 28-05-2012 20:25
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- 28-05-2012 20:36
Unless O is 10m above ground, the s in part iii should not be 5.