Velocity (Magnitude and Direction) of a tossed object as it hits something

[jg871]

Homework Statement

John jumps and shoots and makes a basket from the far end of the court, a distance of 27.5 m. The ball is shot at an angle of 41.0 degrees to the horizontal from a height of 1.98 m above the ground with a speed of 17.2 m/s. The basket is 3.00 m off the ground. a. Sketch a y vs. x graph for the basketball. b. What is the velocity of the basketball (magnitude and direction) as it hits the basket? **I've seen people on yahoo answers try to solve something like this is the conservation of energy, I can't use this.

Homework Equations

The Attempt at a Solution

To be honest, I don't even know where to start. I'm lost and any help would be appreciated[/B]

 

[mfb]

You should be able to make a sketch even without any formal physics knowledge. You can draw the floor, John and the basket, and the approximate path the ball takes.

Concerning the other question, what are the initial vertical and horizontal velocities? How do they change with time?

 

[jg871]

You should be able to make a sketch even without any formal physics knowledge. You can draw the floor, John and the basket, and the approximate path the ball takes.

Concerning the other question, what are the initial vertical and horizontal velocities? How do they change with time?

Sorry, I already got the sketch down I just copy pasted the whole question here. Initial vertical velocity I assume is 17.2 but then again I'm so lost

 

[mfb]

Concerning the other question, what are the initial vertical and horizontal velocities?[...]

That is a very basic question relative to the problem, so it should not be hard to answer this.

 

[HalfLostAndSearching]

I would approach this problem by breaking it down into smaller components and using the principles of physics to analyze each component separately before putting them together to get the final solution.

First, let's consider the motion of the ball in the x-direction. We can use the equation x = x0 + v0x*t + 1/2*a*t^2, where x0 is the initial position, v0x is the initial velocity in the x-direction, a is the acceleration (which is zero in this case since there is no force acting in the x-direction), and t is the time. We know that the initial position is 0 since the ball is shot from the end of the court, and the initial velocity in the x-direction is 17.2*cos(41) m/s. We can also calculate the time it takes for the ball to reach the basket using the equation y = y0 + v0y*t - 1/2*g*t^2, where y0 is the initial position in the y-direction, v0y is the initial velocity in the y-direction, and g is the acceleration due to gravity (9.8 m/s^2). Solving for t, we get t = 1.6 s.

Next, let's consider the motion of the ball in the y-direction. Using the same equation as before, but now with y0 = 1.98 m and v0y = 17.2*sin(41) m/s, we can calculate the height of the ball at any given time. At the time it reaches the basket, the height of the ball will be 3 m.

Now, let's put these two components together to get the final solution. At the time the ball reaches the basket, it will have traveled a horizontal distance of 27.5 m and a vertical distance of 1.02 m (3 m - 1.98 m). Using the Pythagorean theorem, we can calculate the magnitude of the velocity as sqrt((27.5 m)^2 + (1.02 m)^2) = 27.6 m/s. To find the direction of the velocity, we can use the inverse tangent function - tan^-1(1.02/27.5) = 2.1 degrees above the horizontal.

In summary, the velocity of the ball as it hits the basket is 27.6 m/s at an angle of