Question

Basketball player darrell griffith is on record as attaining a standing vertical jump of 1.2 m (4 ft). (this means that he moved upward by 1.2 m after his feet left the floor.) griffith weighs 890 n (200 lb). if the time of the part of the jump before his feet left the floor was 0.300 s, what was the magnitude of his average acceleration while he was pushing against the floor?

Answer 1

Answer:

Magnitude of acceleration = 25.97 $m/s^2$

Explanation:

 We have $v^2=u^2+2as$

 Where v = final velocity, u = initial velocity , s = displacement and a is acceleration

 Here v = 0 m/s, a = acceleration due to gravity =  -9.8$m/s^2$, and s = 1.2 m

So,  $0=u^2-2*9.8*1.2\\ \\ u=\sqrt{2*9.8*1.2} \\ \\ u=4.85m/s$

We know that, Force = Change in momentum by time = m(u-v)/t

So, to jump a height of 1.2 m, we have initial velocity = 4.85 m/s, final velocity = 0 m/s ,mass of person = 890/9.8 = 90.82 kg, and time taken = 0.3 seconds

 Force required = 90.82*(4.85-0)/0.3= 1468.26 N

To lift players body he need to apply a force of mg on the ground = Weight = 890 N

Total force applied =  1468.26+890 = 2358.2 N

Magnitude of acceleration = Force / Mass = 2358.2/90.82 = 25.97 $m/s^2$

Answer 2

Magnitude of player’s average acceleration is $\boxed{25.97\text{ m/s}^2}$.

Further Explanation:

The player follows the Newton’s law of motion while jumping above the ground.

Given:

The vertical distance covered is $1.2\text{ m}$.

Weight of the player is $890\text{ N}$.

Time taken to change the momentum is $0.3\text{ s}$.

Concept:

The mass of the player is given as:

$m=\dfrac{\text{weight}}{\text{acceleration due to gravity}}$

Substitute the values of weight and acceleraton due to gravity.

$\begin{aligned}M&=\dfrac{890}{9.8}\text{ kg}\\&=90.82\text{ kg}\end{aligned}$

To calculate the initial velocity of jump when final velocity, acceleration and distance covered is given, apply the Newton equation of motion given as:

${v^2}={u^2}-2as$

Rearrange the above equation for initial velocity $u$.

${u^2}={v^2}+2as$

Here, $v$ is the final velocity, $u$ is the initial velocity, $s$ is the displacement and $a$ is the acceleration.

Substitute $0\text{ m/s}$ for $v$, $1.2\text{ m}$ for $s$ and $9.81\text{ m/s}^2$ for $a$ in the above equation.

\begin{gathered}

$\begin{aligned}{u^2}&={({0\text{ m/s})^2}+({2\times9.81\text{ m/s}^2\times1.2\,{\text{m}}})\\&=23.544\text{ m}^2/\text{s}^2\\\end{gathered}$

Taking square root on both sides, we get :

$u=4.85\text{ m/s}$

We know that, Force is equal to the rate of change of momentum

Write the equation for relation between force and momentum:

$F=\dfrac{m(u-v)}{t}$

Substitute $4.85\text{ m/s}$ for $u$, $0\text{ m/s}$ for $v$, $0.3\text{ s}$ for $t$ and $90.82\text{ kg}$ for $m$ in above equation.

$\begin{aligned}F&=\frac{{90.82\left( {4.85 - 0} \right)}}{{0.3}}\,{\text{ N}}\\&= 1468}}{\text{.26}}\,{\text{ N}}\\\end{aligned}$

 

To lift the body of player, he needs to apply a force equal to his weight on the ground.

Total force applied by the player on the ground is:

$\begin{aligned}{F_T}&=1468.26\,{\text{ N}} + 890\,{\text{ N}} \\&=2358}}{\text{.26}}\,{\text{ N}}\\\end{aligned}$

Here, ${F_T}$ is the total force applied on the body.

Magnitude of average acceleration:

$\begin{aligned}{a_v}&=\frac{{{F_T}}}{m}\\&=\frac{{2358.26}}{{90.82}}\text{ m/s}^2\\&=25.97\text{ m/s}^2\\\end{aligned}$

 

Thus, the magnitude of player’s average acceleration is $\boxed{25.97\text{ m/s}^2}$.

Learn more:

1. Find the odd one out for lever: https://brainly.com/question/1073452

2. Principle of conservation of momentum: https://brainly.com/question/9484203

3. Motion on the rough surface against friction: https://brainly.com/question/7031524

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

Basketball player, Darrell, Griffith, record, vertical jump, 1.2 m (4 ft), 890 N, feet, 0.300 s, magnitude, average acceleration, the floor, momentum, force, velocity, 200 lb, time.