You throw a ball straight up into the air, releasing it with aspeed of 20 m/s. A

Question

You throw a ball straight up into the air, releasing it with aspeed of 20 m/s. Assuming g = 10m/s2, you catch the ball 4.0 seconds later at the same point from which youlet it go. Consider the motion from just after you release the balluntil just before it returns to your hand. Neglect air resistance,and define up to be positive. Please explainhow to do this. Not just show the answer.. Thanks so much! :)

1)For the round trip what is the ball's average SPEED?
2) For the round trip what is the ball's average ACCELERATION?

Explanation / Answer

Since u(initial velocity)=20m/s we can use the equation v=u+at to solve this question. On the way up: v=u+at a=-10m/s2 (since it is directed downwards) v=0 (since it comes to rest at the top of the path) 0=20-10t t=2s... Therefore it takes 2 seconds to go up and 2 seconds to comedown. speed when it leaves the hand=+20m/s speed when it reaches the hand: v=0+10*2 (since from the top, the velocity is zero) v=20m/s (directed downwards) avg speed=(20+20)/2 =20m/s [NOTE: direction does not apply when we talk about speed as it is ascalar quantity. However, velocity is not] acceleration when moving up: -10m/s2accelerationwhen coming down: 10m/s2 avg acc=(-10+10)/2   --negative sign taken becauseacceleration is a vector quantity.           =0 m/s2

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