Question

a object is projected from ground reaches a point x in its point in its path after 3 seconds and from there it reaches the ground after further 6 seconds the vertical distance of tge point x from the ground is

Answer 1

Answer:

ANSWER

Time of ascent = Time of descent

Therefore the body to reach maximum height will take 2.5 s.Let the maximum height be h and velocity there be v=0.

Using v=u+gt

0=u−9.8×2.5

u=24.5 m/s

Now using work energy theorem u

2

=2gh

h=

2×9.8

24.5×24.5

=30.625 m

Here h is the maximum height reached in 2.5 s,that means point A is in the descending path and will be taking more 1.5 s from the maximum height h. Consider x be the distance of point A from h.Hence height of point A above the ground will be given by,

h−x=

2

1

gt

2

30.625−x=0.5×9.8×1.5×1.5

x=19.6 m