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When throwing a ball straight up when in the air with an initial velocity of 10 meters per second what high will it go and how long will it take to return to the ground?
Wiki User
∙ 11y ago
The initial velocity is 10 meters/sec and is thrown up against the gravitational pull of the earth. This means that the ball is experiencing a deceleration at the rate of 9.8 meters/sec/sec to bring its final velocity to zero.
v^2 - u^2 = 2gs where u is the initial velocity, v the final velocity, g is the acceleration or deceleration, and s is the distance traveled.
0^2 - 10^2 = 2 x (-9.8) x s
-100 = -19.6s
100 = 19.6s
s = 100/19.6 = 5.102 meters
Now v = u + gt where v is the final velocity, u is the initial velocty, g is the acceleration or deceleration, and t is the time. When the ball is thrown up with 10 meters/sec velocity it is acted upon by the deceleration of gravity until its velocity becomes zero.
So 0 = 10 - 9.8t
or 9.8t = 10
t = 1.020 seconds
The time for the ball to go up is 1.020 seconds and the same time is taken for the ball to come back for a total of 2.040 seconds.
Wiki User
∙ 11y ago
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