Question

10. A ball is thrown vertically upwards with an initial velocity of 49 m/s. Calculate the
(i) maximum height attained, (ii) the time taken by it before it reaches the ground
again. (g = 9.8 ms)

Please solve it for me ​

Answer 1

Explanation:

u=49

v=0

v^2-u^2=2gh

0-2401=2×-9.8×h

on solving

we get h=122.5m

v=u+gt

0=49-9.8t

that is time taken=5s

Answer 2

Explanation:

use: v = u -gt (for time)

u= 49 m/s

v = 0

a= 9.8 m/s²

0= 49 -9.8t

On solving: t= 5 sec

total time = 5 + 5 sec = 10 sec

Now use: V² -u² = 2as

0²-49² = 2(-9.8)s

On solving : s= 122.5 m i.e. maximum height achieved by the ball.

Thank You!!