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A ball is thrown vertically into the air with initial velocity 5m/s. Find the time taken to return to the ground and the maximum height achieved.
(1) We assume that the initial height is zero, and that we can neglect air resistance, spin, etc...
The falling object formula is `s=-4.9t^2+v_0t+s_0` where s is the height at time t, s measured in meters, t measured in seconds, `v_0` is the initial velocity in m/s, and `s_0` the initial height in meters.
(2) When the ball hits the ground, s=0 so
`-4.9t^2+5t=0` (Using the convention that velocity towards the earth is negative)
`t(-4.9t+5)=0 => t=0 " or " t=(-5)/(-4.9)~~1.02` . So the time it takes to return to the ground is approximately 1.02 seconds. (The solution of t=0 is disregarded, as the ball is assumed to have started at ground level)
(3) The maximum height is the vertex of the parabola. The time at which the ball assumes its maximum height is found by `t=(-5)/(2(-4.9))~~.51 "seconds"` . The height is :
So the maximum height is approximately 1.28 meters.
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