Question

A ball player catches a ball 3.5 s after throwing it vertically upward.With what speed did he throw it?What height did it reach in m?

Answer 1

Concepts and reason

The concept used to solve this problem is kinematic equations of motion.

Initially, find the speed of the ball after throwing it by using the equation for the velocity as a function of time.

Finally, determine the height the ball reaches by using equation for the displacement as a function of time.

Fundamentals

The expression for the velocity as a function of time is as follows:

$v = u + at$

Here, v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

The expression for the displacement as a function of time is as follows:

$s = ut + \frac{1}{2}a{t^2}$

Here, s is the vertical displacement.

(1)

The speed of ball is,

$v = u + at$

Substitute $\left( { - \;g} \right)$for a

$v = u - gt$

Substitute$\left( { - u} \right)$for v, $\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)$ for g, and 3.5 s for t.

$\begin{array}{c}\\ - u = u - \left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {3.5\;{\rm{s}}} \right)\\\\2u = 34.3\\\\u = 17.2\;{\rm{m/s}}\\\end{array}$

(2)

The height the ball reaches when it is thrown upwards is,

$s = ut + \frac{1}{2}a{t^2}$

Substitute $\left( { - g} \right)$for a and$\left( {\frac{T}{2}} \right)$for t

$\begin{array}{c}\\s = u\left( {\frac{T}{2}} \right) + \frac{1}{2}\left( { - g} \right){\left( {\frac{T}{2}} \right)^2}\\\\ = u\left( {\frac{T}{2}} \right) - \frac{1}{2}g{\left( {\frac{T}{2}} \right)^2}\\\end{array}$

Here, T is the total time taken by the ball to reach the maximum height and to return the ground.

Substitute 17.2 m/s for u, 3.5 s for T, and$\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)$for g.

$\begin{array}{c}\\s = \left( {17.2\;{\rm{m}}} \right)\left( {\frac{{3.5\;{\rm{s}}}}{2}} \right) - \frac{1}{2}\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right){\left( {\frac{{3.5\;{\rm{s}}}}{2}} \right)^2}\\\\ = 30.1 - 15.01\\\\ = 15.09\;{\rm{m}}\\\end{array}$

Ans: Part 1

The speed of the ball after throwing it is 17.2 m/s.

Part 2

The height reached by the ball is 15.09 m.