Question

A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at an angle 55º above the horizontal. What is the impulse delivered by the foot (magnitude and direction)? Ans: 43.6 kg*m/s @ 61.7° above the horizontal

Answer 1

The question revolves around calculating the impulse delivered by a foot to a 2.0 kg ball, which was dropped vertically and then kicked at an angle. The impulse depends upon the change in the ball's momentum after being kicked and can be found by calculating the momentum components before and after the kick and then using vector subtraction to determine the total impulse vector.

Step-by-step explanation:

The student is asking about the impulse delivered by the foot to a ball in a punting situation in Physics. Impulse is defined as the change in momentum of an object when it is subjected to an external force and can be calculated as the product of the average force and the time over which it acts.

To find the impulse, we need to consider the initial and final momentum vectors of the ball. Initially, the ball is dropping vertically and therefore only has vertical momentum when it hits the foot. Using the laws of Physics, we calculate the final vertical and horizontal components of the ball’s velocity after it leaves the foot, multiply them by the mass to get the final momentum, and subtract the initial momentum from it.

The impulse, I, is given by the vector difference between the final and initial momentum vectors:

• Ix = m * Vfx - m * Vix
• Iy = m * Vfy - m * Viy

Where “m” is mass, “Vfx” and “Vfy” are the final horizontal and vertical velocities, and “Vix” and “Viy” are the initial horizontal and vertical velocities. We then find the magnitude and direction using Pythagoras theorem and trigonometry. The impulse will have a magnitude, which is the square root of the sum of the squares of the horizontal and vertical components and a direction given by the arctan of the vertical component divided by the horizontal component.

Answer 2

Step-by-step explanation:

Given

mass of drop
`m=2 kg`

height of fall
`h=1 m`

ball leaves the foot with a speed of 18 m/s at an angle of
`55^(\circ)`

Velocity of ball just before the collision with the floor

`u^=2gh`

`u=√(2gh)`

`u=√(2* 9.8* 1)=4.42 m/s`

Impulse delivered in Y direction

`J_y=m(v\sin (55)-(-u))`

`J_y=2(18\sin (55)+4.42)`

`J_y=38.32 kg-m/s`

Impulse in x direction

`J_x=m* v\cos (55)`

`J_x=2* 4.42\cos (55)=20.646`

`J_(net)=√(J_x^2+J_y^2)`

`J_(net)=√((38.32)^2+(20.64)^2)`

`J_(net)=43.52 kg-m/s`

at an angle of
`\tan \phi =(J_y)/(J_x)=(38.32)/(20.64)`

`\phi =tan^(-1)(1.856)`

`\phi =61.7^(\circ)`