Question

Somone please help me in the 6th question.
Fast its urgent.
Tomorrow is my 10th board exams
Please someone help me.

Answer 1

Given: g=10m/s

Total time=12 sec

1) Total time is 12 sec

Time taken to go up=Time taken to get down

So t=6sec

and at max height v= 0

By equation of motion0=u - gt( - sign for upward direction)

u= 10 (6)

u=58.8 m/s2)

2)At max height v=0 u=58.8t= 6

By 2nd equation of motions=ut- 1/2gt2

s=(58.8×6)-1/2(10)(6×6)

s=352.8- 180s=180m3)

3)Ball reached max height in 6sec

so, in 8sec ball travelled for 2sec after reaching max height which gives u=0t=2sec.

s=?

Again

s=ut + 1/2gt2 ( + sign because ball is coming down)

s=0 + 1/2(10)(2×2)s=20 m