Answer:
Speed of ball at landing: [tex]20\; {\rm m\cdot s^{-1}}[/tex].
Time required: [tex]1\; {\rm s}[/tex].
(Assumption: air resistance on the ball is negligible.)
Explanation:
Assuming that air resistance on the ball is negligible, the ball would accelerate (downward) at a constant [tex]a = g = 10\; {\rm m\cdot s^{-2}}[/tex]. Since acceleration is constant, speed of the ball right before landing can be found using the following SUVAT equation:
[tex]v^{2} - u^{2} = 2\, a\, x[/tex],
Where:
Since the ball was "dropped", the initial velocity of the ball would be [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]. Rearrange the equation and solve for the speed right before landing, [tex]v[/tex]:
[tex]\begin{aligned}v &= \sqrt{u^{2} + 2\, a\, x} \\ &= \sqrt{(0)^{2} + 2\, (10)\, (20)}\; {\rm m\cdot s^{-1}} \\ &= 20\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
In other words, the speed of the ball would be [tex]20\; {\rm m\cdot s^{-1}}[/tex] right before landing.
Acceleration is the rate of change in velocity. To find the duration of the motion, divide the change in velocity by acceleration (which represents the rate of change.)
In this question, the velocity of the ball has changed from the initial value of [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] to [tex]v = 20\; {\rm m\cdot s^{-1}}[/tex] (downward) right before landing. Divide the change in velocity by acceleration to find the duration of the motion:
[tex]\begin{aligned}t &= \frac{v - u}{a} \\ &= \frac{20\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}}{10\; {\rm m\cdot s^{-2}}} \\ &= 2\; {\rm s}\end{aligned}[/tex].
In other words, the ball was in the air for [tex]2\; {\rm s}[/tex].
