if the ball is thrown with an initial velocity of 50 m/s what is the final velocity of the ball at a point at the same level as when it was thrown
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if the ball is thrown with an initial velocity of 50 m/s what is the final velocity of the ball at a point at the same level as when it was thrown
pa help po
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Answer:
Since it doesn’t really matter what the weight of the ball is, we can treat this as a kinematics problem.
For constant acceleration, we have a=(v-vo) /t, so that t= (v-vo) /a where vo is initial velocity.
So the time to peak is 0-(-50)/9.8 m/s^2 = 5.1 seconds, because v, the final velocity at peak, is 0.
Now, of course the trip is not over, and there needs to be whatever went up to come down. Is that the same time span? It would seem logical.
Prove it by the distance formula, p= 1/2 a*t^2+vo* t+po, where po is initial position=0.
Using t=5.1seconds, p=po+t*(v-vo) /2 , so p=t*(50/2)= 127.5 meters.
Now the distance back down is the same 127.5 meters. How long will it take to fall? Now the initial velocity, at the peak, is zero.
The acceleration is 9.8 m/s^2, and the distance is 127.5 m. So 127.5 =½at^2 = ½ 9.8 * t^2.
t=√(255/9.8)=5.1 seconds. So the total trip is 2×5.1=10.2 seconds reaching a height of 127.5 meters.
What goes up comes back down in the same time interval.
Explanation:
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