the ball is dropped from top of the tower, after 2 seconds another ball is thrown vertically downward with a speed 40m/sec after how much time and at what distance the ball meets
Answers
Answer:
If a ball is thrown up from a velocity of 40 m/s, what is the velocity after 6 seconds?
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Using v=u+at first we have to find at what time the ball stops in the air …
Initial velocity=40m/s
Final velocity=0, acceleration=-9.8m/s²(since the ball is thrown upwards)
Using v=u+at
0=40+(-9.8)t
-9.8t=-40
t=40/9.8
t=4.08(let's say it's 4seconds)
Now for the next two seconds the ball will travel downwards
t=2secs
Initial velocity=0
Acceleration=9.8m/s²(travelling downwards)
Final velocity=?
Using v=u+at
V=0+2x9.8
V=19.6m/s downwards
So after 6seconds the ball is travelling at velocity 19.6m/s downwards
Explanation:
Velocity after time t for upward motion is v=u-gt. Velocity v will be zero after 40/9.8 or 200/49 seconds. After that time the ball will start falling down and the velocity after 6 seconds when the ball was thrown up will be 9.8*(6–200/49) which comes out to be 18.8 m/s. This final velocity will be downwards.
Answer:
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