Question

A ball is tossed vertically upward with an initial speed of 27.3 m/s. How long does it take before the ball is back on the ground?

Answer 1

The ball will go up to a certain height, and then it will start to fall back to the ground until it hits it and stop moving. You could use the kinematics equation for accelerated motion in vertical direction:
\[Y_{f} = Y_{o} + V_{o_{y}} * t + \frac{ 1 }{ 2 } * a * t^2\]
Where:
\[Y_{o}=Y_{f} = 0m\]
\[V_{o_{y}} = 27.3 \frac{ m }{ s }\]
\[a = -g = -9.8 \frac{ m }{ s^{2} }\]

Then you are only left with:
\[0 = 0 + V_{o_{y}} * t - \frac{ g*t^{2} }{ 2 }\]
\[\frac{ g*t^{2} }{ 2 } = V_{o_{y}} * t\]
Which gives either:
\[t_{1} = 0s\]
or
\[\frac{ g*{t_{2}}^{2} }{ 2 } = V_{o_{y}} * t_{2}\]
\[\frac{ g*t_{2} }{ 2 } = V_{o_{y}}\]
\[t_{2} = \frac{ 2*V_{o_{y}} }{ g }\]
\[t_{2} = \frac{ 2*27.3 }{ 9.8 } s\]
\[t_{2} = 5.57 s\]

And getting two results is nothing crazy, they are just saying that the ball is at the ground (y=0m) at two instants of time t. At time t=0s (When you first toss the ball), and at time t=5.57s (When it hits the ground after falling).

Answer 2

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