Question

A ball is thrown up vertically returns - to the thrower after 10 seconds.

Calculate :-
(a) the Velocity with which it was thrown up.
(b) the maximum height it reaches
(c) its position after 7 second.​

Answer 1

Answer:

Explanation:g=10(up) , g= -10(down)

given: t=10s

1)t=2u/g

10=2u/10

u=50m/s

2)H(max)=u2/2g

=2500/2 x 10

H=125m

3)t=7s

s=ut+1/2at2

=50 x 7+1/2 (-10) 7 x 7

350 -5 x 49

350 - 245 = 695m

may it helps and if please say thanks and brainly

Answer 2

Explanation:

here assume that time of ascent is equal to the time of descent therefore the total time to cover the height is 10÷2=5 so then solve. now after the ball will complete its first displacement in 5 sec its final velocity will be zero and also in the 3rd bit is has been asked that the position of the ball after 7 sec which means as the ball completed its first displacement in 5 sec now from the top where the final velocity was 0 now calculate the displacement after 2 second why because 5 +2=7 then subtract the displacement in 2 sec from the displacement in 5 sec the difference will give u the position of the ball after 7 sec or u can say that the ball's position above the ground thankyou