Question

A body projected up reaches a point A in its path at the end of 4th second and reaches the sound
after 5 seconds from the start. The height of A above the ground is (g = 10m/s)​

Answer 1

Answer:

Time of ascent=Time of descent=2.5s as total time is 5s

Let x be the height of point A

At max. height, v=0. Thus,

v=u+at

0=u+(-g)t

u=gt and thus u=(10)2.5= 25m/s

WKT, u2=2gH

H=25(25)/20=31.25m

We can infer that point A is in the path of descent.

Now starting from the max height H, u=0. Already 2.5 seconds are over. So to reach A the body must travel 1.5 seconds more. If H-x is the distance of A from max height,

H-x=1/2gt2

H-x=1/2(10)(1.5)2

H-x=2.25(5)=11.25

x=31.25-11.25

Thus, x=20m

Final answer=20m

Explanation:

help you

Answer 2

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