Question

A ball is thrown up with a speed of 4.5 m/s.
(i) How high will it go before it begins to fall?
(ii) How long will it take to reach that height?

Answer 1

Answer:

A   given :-

 u = 0.5 m/s

 g = 9.8 m/s^2 ( it is taken to be negative since body is going against the gravity .

 v = 0

 s = ?

v^2 - u^2 = 2gs  

(0)^2 - (0.5)^2 = 2(-9.8) S

- (0.5 * 0.5 ) =  -2  * 9.8 * S

            S = 0.5 *0.5 / 2 * 9.8 = 0.012 m is answer

B   V=u+gt

V=0m/a

U=0.5m/s

g=-10

0=0.5-(10.t)

-0.5=-10t

t=0.05

Pls mark as the brainliest answer

Answer 2

Answer:

Final speed at the top most point (v) = 0

Initial speed (u) = 15 m/s

a = -g = -9.8

Hence, as per the third law of motion,/

v square = u square + 2.a.s

or distance s = (v square - u square)/2.a

or s = (0 - 225)/(2 x -9.8)

or s = 225/19.6 = 11.48 mtrs Answer