Find the distance travelled by ball in 5 sec when it is thrown in upload direction with
speed 40 m/s
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Answer:
10m distance .. because I'm in class 7 and I don't know lol
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Answer:- According to the equation of the motion under gravity
v
2
−u
2
=2gs
u=initial velocity of the ball=40m/s
v= Final velocity of the ball=0m/s
Let h be the maximum height attained by the ball
Therefore,
0
2
−40
2
=2(−10)h
h=(40×40)/20=80
Therefore, total distance covered by the stone during its upward and downward journey=80+80=160m
Net displacement during its upward and downward journey=80+(−80)=0
Explanation:
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