Question

Q.13. A ball is thrown vertically upwards with a velocity
of 49 m s-1. Calculate :
(i) the maximum height to which it rises,
(ii) the total time it takes to return to the surface of
the earth.Explain step by step​

Answer 1

Solution:

Given:

$\bullet$ v = 0 m/s

$\bullet$ u = 49 m/s

$\bullet$ Upward motion (g) = - 9.8 m/s^-2

Now:

Let h be the maximum height attained by the ball.

Using:

$\boxed{\sf{v^{2} - u^{2} = 2h}}$

We get:

$\implies$ 0^2 − 49^2 = 2 (−9.8) ℎ

$\implies$ ℎ = 49 × 49 / 2 × 9.8

$\implies$ ℎ = 122.5

Now:

Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:

$\boxed{\sf{v = u + gt}}$

So:

$\implies$ 0 = 49 + (−9.8) t

$\implies$ 9.8t = 49

$\implies$ t = 49/9.8

$\implies$ t = 5 s

But,

Time of ascent = Time of descent

Therefore,

Total time taken by the ball to return:

$\implies$ 5 + 5

$\implies$ 10 s

______________________

Answered by: Niki Swar, Goa❤️

Answer 2

Best of luck yr jehda v exam aaa

my best wishes are always with you ❤️