An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . how fast wil

Question

Vesna [10]

4 years ago

11

An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . how fast wil

l he be moving backward just after releasing the ball?

Physics

2 answers:

Mamont248 [21] · Answer 1 · 2021-11-28T22:51:30+03:00

Mamont248 [21]4 years ago

6 0

Applying conservation of momentum
Quarterback mass = 80 kg
ball mass = 0.43 kg
Initially both together but horizontal velocity of both 0
initial momentum = 0
Final momentum = 15*0.43 - 80v
initial = final (law of conservation of momentum)
6.45 = 80v
v = 0.08 m/s

geniusboy [140] · Answer 2 · 2021-11-28T23:56:05+03:00

The speed of the quarterback after throwing the football horizontally is $\boxed{0.08\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ .

Further Explanation:

Given:

The mass of the quarterback is $80\,{\text{kg}}$ .

The mass of the football is $0.43\,{\text{kg}}$ .

The speed of the ball in the forward direction is $15\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ .

Concept:

The quarterback jumps upward and throws the football horizontally at the speed of $15\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ . As the quarterback throws the ball in forward direction horizontally, the ball gains some momentum and as the ball gains momentum in the forward direction, there will be some momentum gained by the quarterback in the backward direction due to conservation of momentum.

The initial momentum of the system in the horizontal direction is zero as both have no velocity in the horizontal direction. Therefore, by the conservation of momentum, the Quarterback will gain the same momentum as that of ball but in the opposite direction so that the final momentum of the system in the horizontal direction remains zero.

By applying the conservation of momentum in horizontal direction:

$\boxed{{M_m}\times{V_m}={M_b}\times{V_b}}$

Here, ${M_m}$ is the mass of the person, ${V_m}$ is the speed of the person in backward direction, ${M_b}$ is the mass of the football and ${V_b}$ is the horizontal speed of the football.

Substitute the values in the above expression.

$\begin{aligned}80\times{V_m}&=0.43\,{\text{kg}}\times{\text{15}}\\{V_m}&=\frac{{6.45}}{{80}}\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\&=0.080\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

Thus, the speed of the quarterback after throwing the ball horizontally will be $\boxed{0.080\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ .