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If 4sec be the time in which a projectile reaches a point P of its path and 5 sec the time from P till it reasches the horizontal plane passing through the point of projection . The height of P above the horizontal plane(in m) will be:-????? [g=9.8m/sec 2 ]

If 4sec be the time in which a projectile reaches a point P of its path and 5 sec the time from P till it reasches the horizontal plane passing through  the point of projection . The height of P above the horizontal plane(in m) will be:-?????
[g=9.8m/sec2]

Grade:12

4 Answers

bharath
31 Points
7 years ago
v=u+at       as v=u+at we can abe to find v as u=zeo   and h=vsquare /2g   from it we can find h
v=36 and t in that particular projectile is 4 sec so h=1296/18
h=72
Hansraj Gyanendra Singh Rajawat
35 Points
7 years ago
no this answer is not correct you have made a mistake yhe answer of this question is 98 plz check again then send the right answer...
Aditya Subbaraman
21 Points
7 years ago
initially the total time of flight of projectile 2usin0/g =9.Then from there we will get the value of usin0.Take the movement about the vertical direction.s=ut+1/2at^2.where u is usin0 and a is g.From there when you do the calculations you get the answer as 98.
Hopefully this answers your query
Suryanshu Tiwari
15 Points
5 years ago
Since, we need to calculate the height at 'P' after 4 seconds. So what you have to do just simply put 4 in place of 'T' in the formula of Time of flight from where you will get the value of usin ₩. Then put it in S = usin₩ t + 1/2 g t^2. Where t=4sec. 
That's all

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