A person throws some balls into the air after every second. The next ball is thrown when the velocity of first ball is zero. How high do the balls rise above his hand.
Answers
Explanation:
Given A person throws some balls into the air after every second. The next ball is thrown when the velocity of first ball is zero. How high do the balls rise above his hand .
- A person throws a ball into the air one after the other. When the ball is thrown up, the ball stops, so the final velocity is zero and time taken is 1 sec. We need to find the height.
- So we have equation of motion
- So v = u + at but a = - g
- So v = u – gt
- 0 = u - 10 x 1
- Or u = 10 m / s
- Now initial velocity u = 10 m/s
- So we have v^2 = u^2 – 2gh
- 0 = (10)^2 – 2 x 10 x h
- 0 = 100 – 20 h
- 20 h = 100
- Or h = 100 / 20
- Or h = 5 m
- So the ball will rise 5 m above his hand.
Reference link will be
https://brainly.in/question/6927452
☆The person throws balls into air vertically upward in regular intervals of time of one second.And next ball is thrown when the velocity of the ball thrown earlier becomes zero.
☆The person throws balls into air vertically upward in regular intervals of time of one second.And next ball is thrown when the velocity of the ball thrown earlier becomes zero.☆So time to reach the highest point is 1 sec. If initial velocity is u then at highest point,
☆given :-
0=u−gt
t=1s g=10m/s2
u=10m/s
☆Height (H) to which ball rises ,
0=u2-2gH
H=2gu2=5m
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