Question

A ball is thrown vertically upward from the ground. It crosses a point at the height of 25 m twice at an interval of 4 seconds. The ball was thrown with the velocity of
A) 18 m/s
B) 25 m/s
C) 30 m/s
D) 36 m/s

Answer 1

$\huge\orange{ASSALAMUALAIKUM!}$

After reaching the maximum height, it falls for
2 second to reach the 25m height.

distance it has fallen is-

h=0.5gt^

0.5*10*4 =20m

so the maximum height- 25+20 =45m

then velocity at the bottom or the initial

velocity is found from

u^=2gh

=2*10*45=900

u=30m/s

$\huge\orange{INSHAALLAH it will help you!}$

Answer 2

Answer:AT top

0  =  u – at

0 = u -10*2

u = 20 m/s (t = 2s)

u becomes final velocity for the ground projection thus,

400 = u^2 – 2*-10*25

u = 30 m/s

This would be gthe velocity.

Explanation: