It travels upwards until the vertical velocity component becomes zero. An easier way to calculate it it to require that the initial kinetic energy become completely potential energy there.
(1/2) M Vo^2 = M g H
where H is the maximum height above the thrower's hand.
H = Vo^2/(2 g) = 11.5 m
The time in the air is twice the time it takes for the velocity upward to become zero. It spends an equal length of time coming back down. Therefore:
T = 2 * (Vo/g) = 3.06 s
-A person throws a ball upward into the air with an initial velocity of 15 m per second.
- How could you calculate how high the ball goes and how long the ball is in the air before it comes to the thrower's hand?
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thanx so much! im finally starting to get this!
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