a ball is thrown up with a speed of 15 m per sec how high will it go before it begins to fall
Answers
Answered by
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
Similar questions