A ball is tossed with enough speed straight up so that it

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(a)

When a ball is thrown vertically upward, its velocity decreases due to the acceleration of gravity until it reaches its highest point. At the highest point, the velocity becomes zero before it starts to descend. This is because at the highest point, the ball momentarily comes to a stop before gravity starts pulling it back down.

\(v = 0\)

(b)

The ball’s velocity \(1\;{\rm{s}}\) before it reaches its highest point is,

\(v = {v_1} + at\)

\(0 = {v_1} + \left( { - g} \right)t\)

\({v_1} = gt\)

Here, \(g\) is the acceleration due to gravity having an approximately value of \(10\;\frac{{\rm{m}}}{{{{\rm{s}}^2}}}\).

Plug the value of \(10\;\frac{{\rm{m}}}{{{{\rm{s}}^2}}}\) for \(g\), and \(1\;{\rm{s}}\) for \(t\) in the above equation.

\({v_1} = 10 \times 1\)

\( = 10\;\frac{{\rm{m}}}{{\rm{s}}}\)