Question

a ball is thrown up with a speed of 15 m per sec how high will it go before it begins to fall​

Answer 1

Answer:

11.48 mtrs

Step-by-step explanation:

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

hope it helps