Question

An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15. How fast will he be moving backward just after releasing the ball?
Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far will he move horizontally, assuming his speed is constant?

• The quarterback jumping off of the ground. 

• The quarterback throwing the ball. 

• The ball encountering air resistance after being thrown. 

• The quarterback landing back on the ground.

Answer 1

Concepts and reason

The main concepts used to solve the problem are conservation of momentum and relation between distance, speed, and time.

Initially, use the conservation of momentum and use the respective values to get the speed of quarterback backwards. Later, use the value of speed found in the previous step. Finally, use the relation between, distance, speed, and time to calculate the distance.

Fundamentals

The conservation of momentum states that in the absence of external force the momentum remains conserved. It is expressed as follows:

${m_1}{v_1} = {m_2}{v_2}$

Here, ${m_1}$ is the initial mass, ${m_2}$ is the final mass, ${v_1}$ is the initial velocity, and ${v_2}$ is the final velocity.

The relation between distance, speed, and time is,

$s = \frac{d}{t}$

Here, s is the speed, d is the distance, and t is the time.

Calculate the speed of quarterback backwards.

The conservation of momentum states that the initial momentum is equal to the final momentum. It is expressed as follows:

${m_1}{v_1} = {m_2}{v_2}$

Here, ${m_1}$ is the initial mass, ${m_2}$ is the final mass, ${v_1}$ is the initial velocity, and ${v_2}$ is the final velocity.

Rewrite the expression ${m_1}{v_1} = {m_2}{v_2}$ for ${v_1}$ .

${v_1} = \frac{{{m_2}{v_2}}}{{{m_1}}}$

Substitute 80 kg for ${m_2}$ , $15\,{\rm{m/s}}$ for ${v_2}$ , and $0.43\,{\rm{kg}}$ for ${m_1}$ in expression ${v_1} = \frac{{{m_2}{v_2}}}{{{m_1}}}$ .

$\begin{array}{c}\\{v_1} = \frac{{\left( {0.43\,{\rm{kg}}} \right)\left( {15\,{\rm{m/s}}} \right)}}{{\left( {80\,{\rm{kg}}} \right)}}\\\\ = 0.080625\,{\rm{m/s}}\\\end{array}$

Calculate the horizontal distance.

The relation between distance, speed, and time is,

$s = \frac{d}{t}$

Here, s is the speed, d is the distance, and t is the time.

Rewrite the equation $s = \frac{d}{t}$ for d.

$d = st$

Substitute 0.30 s for t and $0.080625\,{\rm{m/s}}$ for s in expression $d = st$ .

$\begin{array}{c}\\d = \left( {0.080625\,{\rm{m/s}}} \right)\left( {0.30\,{\rm{s}}} \right)\\\\ = 0.0241875\,{\rm{m}}\\\end{array}$

Ans:

The speed of quarterback backwards is $8.06 \times {10^{ - 2}}\,{\rm{m/s}}$ .

The distance moved horizontally is $2.42\,{\rm{cm}}$ .

Answer 2

Apply conservation of linear momentum. \(m_{b}\left(v_{\psi f}-v_{b i}\right)=m_{q}\left(v_{q}\right)\)

$$ v_{q}=\frac{m_{b}\left(v_{b f}-v_{b i}\right)}{m_{q}}=\frac{(0.43 \mathrm{~kg})(15 \mathrm{~m} / \mathrm{s}-0)}{80 \mathrm{~kg}}=0.0806 \mathrm{~m} / \mathrm{s} $$

According to the conservation of linear momentum, when the quarterback throws the ball forward, he recoils in back ward direction with same momentum. So, to analyze the conservation of momentum, the interaction 'the quarter back throwing the ball' is to be considered.