Question

A ball is thrown vertically upwards and returns to the thrower after 12 sec (g=9.8 m/s2). Find
1) the velocity with which it was thrown up.
2) the maximum height it reaches.
3) its position after 8 sec

Answer 1

Given: g=9.8
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 motion
0=u - gt( - sign for upward direction)
u= 9.8 (6)
u=58.8 m/s
2) At max height v=0
u=58.8
t= 6
By 2nd equation of motion
s=ut- 1/2gt2
s=(58.8×6)-1/2(9.8)(6×6)
s=352.8- 176.4
s=176.4m
3) Ball reached max height in 6sec
so, in 8sec ball travelled for 2sec after reaching max height
which gives u=0
t=2sec. s=?
Again s=ut + 1/2gt2 ( + sign because ball is coming down)
s=0 + 1/2(9.8)(2×2)
s=19.6 m

Answer 2

Explanation:

if the answer was correct then its op