Impulse basketball homework Question

[creative_wind]

Hello,
I am (obviously) new to the forum. I am in an introductory-level physics course in university and although I had taken a year of physics in high school, these easy intro questions on my homework keep snagging me. This is online homework and I get 5 attempts at the correct solutions. This is what I have:

Homework Statement

To make a bounce pass, a player throws a 0.60 kg basketball toward the floor. The ball hits the floor with a speed of 6.5 m/s at an angle of 64° from the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the floor?

Homework Equations

I=Favg(T2-T1)=P=m*v

Where P=momentum

The Attempt at a Solution

I=m*v
m=0.6kg
v=6.5*cos(64)=2.85m/s
I=0.6kg*2.85m/s=1.71kg*m/s <--this was wrong

I=m*v
m=0.6*cos(64)=0.26kg
v=6.5*cos(64)=2.85m/s
I=0.26kg*2.85m/s=0.74kg*m/s <--this was wrong

I figured the velocity had to be in the y-component since it is vertically upward from the floor. I'm really not sure though...

Thanks :)

 

[physixguru]

this should help ya:

Impulse is the integral of force over time, it is measured in Newton-seconds. For instance a force of one Newton applied over one second will change the momentum, a force of two Newton's applied over half of a second will have a similar effect.

For rigid body collisions we take this to its limit and apply an infinite force over an infinitesimally small time. This impulse is equal to the change in momentum of the colliding objects. Because we are talking about forces here, Newton's third law applies, and the impulse on the colliding objects will be equal and opposite.

impulse = m(vf- vi)

 

[Moetasim]

Hello,

I would like to offer some guidance on how to approach this problem. Firstly, it is important to understand the concept of impulse in physics. Impulse is defined as the change in momentum of an object over a period of time. It is calculated by multiplying the average force acting on an object by the time it is applied. In this case, we can use the equation I = Favg * Δt.

Now, let's look at the given information. We know that the basketball has a mass of 0.60 kg and a velocity of 6.5 m/s at an angle of 64° from the vertical. We also know that after hitting the floor, it rebounds with the same speed and angle. This means that the change in momentum (Δp) of the basketball is equal to its initial momentum (p) since it ends up with the same speed and direction. Therefore, we can use the equation Δp = p = m * v.

To calculate the impulse, we need to find the average force (Favg) and the time (Δt). We can find the average force by using the equation Favg = Δp/Δt. Since the change in momentum is equal to the initial momentum, we can rewrite this equation as Favg = p/Δt. Now, we need to find the time it takes for the basketball to hit the floor and rebound. This can be calculated by dividing the total distance traveled by the average speed, which is given by v * cos(64°). This gives us Δt = 2h/(v * cos(64°)) where h is the height at which the basketball is thrown.

Now, we can plug these values into the equation for impulse and calculate the answer. I would suggest reviewing the concept of impulse and understanding the equations before attempting the problem again. Remember to pay attention to the units and use the correct formula. I hope this helps!