Question

23 )A ball is thrown vertically upwards with an initial velocity of 49 m/s. Calculate the maximum height attained and time taken by it before it
reaches the ground again. (g = 9.8 m/s2 )

i. 12.25 m, 10s
ii. 122.5 m, 5s
iii. 122.5 m, 10s
iv. 12.25, 5s​

Answer 1

Answer:

u = 49 m/s

g = - 9.8 m/s² , it is negative because it is acting in the upward direction

v = 0 m/s

1. Time taken = v = u-gt

= 0 = 49-9.8t

= 9.8t = 49

= t = 5 seconds

2. Distance Covered = ut-0.5gt²

= (49*5) - (0.5*9.8*25) m

= 245-122.5 m

= 122.5 m

So, option ii is correct.

I hope this helps.