Respuesta :
We have [tex]v^2 = u^2 + 2 a s[/tex]
Where v = final velocity, u = initial velocity , s = displacement and a is acceleration
Here v = 0 m/s, a = -9.8m/[tex]s^{2}[/tex], and s = 0.9 m
So, [tex]0 = u^2 - 2 *9.81*0.9[/tex]
[tex]u = \sqrt{2*9.81*0.9} =4.20 m/s[/tex]
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
The total force required to lift player body is [tex]\fbox{\begin\\2143.54\text{ N}\end{minispace}}[/tex].
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 [tex]0.90\text{ m}[/tex].
The player jump interval is [tex]0.3\text{ s}[/tex].
The mass of the player is [tex]90\text{ kg}[/tex].
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 [tex]0.90\text{ m}[/tex] and time of jump is [tex]0.3\text{ s}[/tex], the term g is called acceleration due to gravity and it has a constant value of [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] and it is taken negative.
The expression for the motion can be written as:
[tex]\fbox{\begin\\{{v^2}=\d{u^2} + 2as}\end{minispace}}[/tex]
Here, [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity, [tex]a[/tex] is the acceleration of the body and [tex]s[/tex] is the distance covered by the body.
Substitute 0 for [tex]v[/tex], [tex]0.90\text{ m}[/tex] for [tex]s[/tex], [tex]0.3\text{ s}[/tex] for [tex]t[/tex] and [tex]- 9.81{\text{ m/}}{{\text{s}}^2}[/tex] for [tex]a[/tex] in the above equation.
[tex]\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}[/tex]
The expression for the momentum is:
[tex]\fbox{\begin\\{F=\dfrac{{m}{(u-v)}}{t}}\end{minispace}}[/tex]
Here, [tex]F[/tex] is the force momentum and [tex]m[/tex] is the mass of the object.
Substitute 0 for [tex]v[/tex], [tex]4.20\text{ m/s}[/tex] for [tex]u[/tex], [tex]0.3\text{ s}[/tex] for [tex]t[/tex] and [tex]90\text{ kg}[/tex] for [tex]m[/tex] in the above equation.
[tex]\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}[/tex]
The weight of the payer is:
[tex]\begin{aligned}W&=\left( {90{\text{ kg}}}\right)\left( {9.81{\text{ m/}}{{\text{s}}^2}}\right)\\&=882.9{\text{ N}}\\\end{aligned}[/tex]
The total force required to lift player body is:
[tex]\begin{aligned}{F_t}&=1260.64{\text{ N}} + 882.9{\text{ N}}\\ &=2143.54{\text{ N}} \\\end{aligned}[/tex]
Therefore, the total force required to lift player body is [tex]\fbox{\begin\\2143.54\text{ N}\end{minispace}}[/tex].
Learn more:
1. The acceleration of the body. https://brainly.com/question/7031524
2. The force acting on the refrigerator. https://brainly.com/question/4033012
3. How will the ball's behavior differ from its behavior on earth https://brainly.com/question/10934170
Answer Details:
Grade: College
Subject: Physics
Chapter: Kinematics
Keywords:
Basketball, player, jump, vertically, [tex]0.9\text{ m}[/tex], assume, jump travel, lasts [tex]0.3\text{ s}[/tex], mass, [tex]90\text{ kg}[/tex], ignore, resistance, required, lift
.
