Question

A man throws a ball into the air one after the other, throwing one when the other is at the highest point. How high do the balls rise if the throws twice a second?
(A) $0.49\;{\rm{m}}$
(B) $1.25\;{\rm{m}}$
(C) $2.45\;{\rm{m}}$
(D) $4.9\;{\rm{m}}$

Answer 1

First, determine the speed of the ball at which it is thrown in the air with the first equation of motion. After this, we can determine the height of the ball from the third equation of motion. Two balls are thrown in the air in one second, so use this information to determine time after which ball reaches the maximum heights.

Complete step by step answer:
If a man throws the ball twice in one second, it means that the ball will take to reach the maximum height.
Write the first equation of motion to calculate the ball's speed at which it is thrown in the air.
$v = u - gt$
Here, $v$ is the speed of the ball at the maximum height, which is equal to zero, $u$ is the speed of the ball at which it is thrown, and $g$ is the acceleration due to gravity (it is negative because the direction of the ball is upward) and $t$ is the time after which ball will reach the maximum height.
Put $g = 10\;{\rm{m}}{{\rm{s}}^{ - 2}}$, $t = \;{\rm{0}}{\rm{.5}}\;{\rm{s}}$ and $v = 0\;{\rm{m}}{{\rm{s}}^{{\rm{ - 1}}}}$ in the above equation.
Therefore, we get
0 = u - \left( {10\;{\rm{m}}{{\rm{s}}^{ - 2}} \times 0.5\;{\rm{s}}} \right)\\\
$\implies u = 5\;{\rm{m}}{{\rm{s}}^{ - 1}}$
After getting the value of $u$, we can determine the maximum height of the ball, so write the third equation of motion.
${v^2} = {u^2} - 2gh$
Here, $h$ is the maximum height of the ball.
Substitute the value in the above equation.
0 = {\left( {5\;{\rm{m}}{{\rm{s}}^{ - 1}}} \right)^2} - \left( {2 \times 10\;{\rm{m}}{{\rm{s}}^{ - 2}} \times h} \right)\\\
$\implies h = \dfrac{{25\;{{\rm{m}}^2}{s^{ - 2}}}}{{20\;{\rm{m}}{{\rm{s}}^{ - 2}}}}\\\$
$\implies h = 1.25\;{\rm{m}}$
Therefore, the ball will rise up to $1.25\;{\rm{m}}$ if a man throws it twice a second

So, the correct answer is “Option B”.

Note:
Remember the equations of motion for calculating terms like velocity, acceleration, time, and many more, because equations relate these terms with each other. Here also we use two equations of motion for the determination of height and speed of the ball.