Question

a ball thrown upward is return back to the ground in 10 seconds.Find the position of the ball after 7 second. (g= 10m/s^2).

Answer 1

Time to reach Maximum height ,
t = 10/2 = 5 s.

v = 0 (at the maximum height )

a = - g =10 m s-²

Using, v = u + at, we get

0 = u - 10× 5

or, u = 50 ms-¹

Thus, the velocity with which it was thrown up = 50 ms-¹

Using, 2aS = v² - u², we get

S = v²- u²/2a

= 0 - (50× 50/- 2× 10)

= 125 m

Thus, Maximum height it reaches = 125 m.

Here, t = 7s. In 5 s, the ball reaches the maximum height and in 2s it falls from the top.

Distance covered in 2 s from maximum height,

S = ut + 1/2at ²

= 0 + 1/2 × 10× 2^2

= 20 m

so the ball is 125-20=105m above from ground.