Physics, 11.07.2019 08:00 emilylunaa

Abasketball player can jump vertically 0.9 m. assume that the jump interval (from lowering the body to his feet taking off) lasts for 0.3 s and the mass of the player is 90 kg. ignore air resistance.

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Answer from: jrfranckowiak

We have v^2 = u^2 + 2 a s

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

Here v = 0 m/s, a = -9.8m/ $s^{2}$ , and s = 0.9 m

So, 0 = u^2 - 2 *9.81*0.9

$u = \sqrt{2*9.81*0.9} =4.20 m/s$

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

So, to jump a height of 0.9 m

Force required = 90*(4.20-0)/0.3s= 1260N

To lift players body he need to apply a force of mg on the ground = 98*9.81 = 883N

Total force applied = 1260+883 = 2143 N

Answer from: Homepage10

The total force required to lift player body is $\fbox{\begin\\2143.54\text{ N}\end{minispace}}$ .

Further Explanation:

The momentum is the vector quantity has some magnitude and unique direction represented in the coordinates system on the x, y and z-axis. It is the related directly to the object mass and object velocity.

Velocity is the vector quantity has some magnitude and some direction represented in the coordinates system on the x, y and z-axis.

Given:

The payer can jump a vertical distance of $0.90\text{ m}$ .

The player jump interval is $0.3\text{ s}$ .

The mass of the player is $90\text{ kg}$ .

Concept:

The player jump from the ground the final velocity is equal to zero.

As we have the final velocity zero, the vertical jump distance of $0.90\text{ m}$ and time of jump is $0.3\text{ s}$ , the term g is called acceleration due to gravity and it has a constant value of $9.81{\text{ m/}}{{\text{s}}^2}$ and it is taken negative.

The expression for the motion can be written as:

$\fbox{\begin\\{{v^2}=\d{u^2} + 2as}\end{minispace}}$

Here, is the final velocity, is the initial velocity, is the acceleration of the body and is the distance covered by the body.

Substitute 0 for , $0.90\text{ m}$ for , $0.3\text{ s}$ for and $- 9.81{\text{ m/}}{{\text{s}}^2}$ for in the above equation.

$\begin{aligned} {0^2}&={u^2}+2\left({ - 9.81{\text{ m/}}{{\text{s}}^2}}\right)\left( {0.9{\text{ m}}}\right)\\{0^2}&={u^2} - 17.658{\text{ }}{\left( {{\text{m/s}}} \right)^2}\\{u^2} &= 17.658{\text{ }}{\left({{\text{m/s}}}\right)^2}\\u&=\sqrt {17.658{\text{ }}{{\left( {{\text{m/s}}}\right)}^2}}\\&=4.20{\text{ m/s}}\\\end{aligned}$

The expression for the momentum is:

$\fbox{\begin\\{F=\dfrac{{m}{(u-v)}}{t}}\end{minispace}}$

Here, is the force momentum and is the mass of the object.

Substitute 0 for , $4.20\text{ m/s}$ for , $0.3\text{ s}$ for and $90\text{ kg}$ for in the above equation.

$\begin{aligned}[tex]F&=\frac{{\left( {90{\text{ kg}}} \right)\left( {4.20 - 0}\right){\text{m/s}}}}{{0.3{\text{ s}}}}\\&=1260.64{\text{ N}}\\\end{aligned}$

The weight of the payer is:

$\begin{aligned}W&=\left( {90{\text{ kg}}}\right)\left( {9.81{\text{ m/}}{{\text{s}}^2}}\right)\\&=882.9{\text{ N}}\\\end{aligned}$

The total force required to lift player body is:

$\begin{aligned}{F_t}&=1260.64{\text{ N}} + 882.9{\text{ N}}\\ &=2143.54{\text{ N}} \\\end{aligned}$

Therefore, the total force required to lift player body is $\fbox{\begin\\2143.54\text{ N}\end{minispace}}$ .

Learn more:

1. The acceleration of the body.

2. The force acting on the refrigerator.

3. How will the ball's behavior differ from its behavior on earth

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Basketball, player, jump, vertically, $0.9\text{ m}$ , assume, jump travel, lasts $0.3\text{ s}$ , mass, $90\text{ kg}$ , ignore, resistance, required, lift