Question

A body projected up reaches a point A in its
path at the end of 4th second and reaches the
ground after 5 seconds from the start. The
height of A above the ground is (g = 10 m/s)
1) 19.6m 2) 30.6m 3) 11m 4) 20m
Here why the time of accent is equal to time of descent​

Answer 1

Explanation:

CALCULATOR

Enter the know values to find unknown

Acceleration

Distance

Final Velocity

v = u + a*t

Initial Velocity u

m/s

Final Velocity v

m/s

Time t

s

Acceleration a

m/s²

RESET VALUES

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