A student shot a basketball from the edge of the top of the stadium into a baske

Question

A student shot a basketball from the edge of the top of the stadium into a basketball hoop on the football field. They threw the ball at an angle of 20 degrees above the horizontal with a speed of 20.0 m/s and that the top of the stadium is 40m above the hoop.
(a) what are the x and y components of the initial velocity?
(b) what is the time that elapses from the moment the ball is released to the moment it goes in the hoop.
(c) what is the x displacement once the ball lands in the hoop?
(d) What are the x and the y components of the final velocity of the ball going through the hoop?
(e)what is the final velocity of the ball? Include magnitude and direction relative to the positive x-axis.

Explanation / Answer

a) You can imagine a right-triangle with angle 20 degrees. The hypotenuse is equal to 20. You can use trig to then find the other two sides, which are the x-component and the y-component. Remember trig? It's sin = opp/hyp and cos = adj/hyp. So, rearranging,

opp = hyp*sin = 20*sin20 = 6.84. This is the y-component.

adj = hyp*cos = 20*cos20 = 18.79. This is the x-component.

b) This is a bit tricky. We want the ball to fall into the basket. But we don't really have a horizontal distance for it to travel... We only have the vertical distance for it to fall, which is 40 m. There's also an intial velocity upward, 6.84 m/s. And then, of course, the acceleration due to gravity downward, usually considered to be 9.8 m/s^2.

Ok, so there's an equation you should have maybe in your book. It should look something like x = v0t + (1/2)at^2. But we will write it like

y = v0yt + (1/2)ayt^2. Here, y is going to be 40 m, v0y is going to be 6.84 m/s, and ay is going to be 9.8 m/s^2. So, 0 = 40 m + (6.84 m/s)t + (1/2)*(-9.8 m/s^2)t^2. Now you have what looks like a quadratic equation. These are solved by the formula that looks like this

x = [-b ± (b^2 - 4*a*c)]/2*a . But here, our equations use t instead of x. So, just pretend that c is 40, b is 6.84, and a is -9.8. Then fill it into this equation and see what happens.

t = 2.399 s. You might want to check my math...

c) Well, the x-component of the velocity of the ball is 18.79 m/s. And the ball takes 2.399 s to land in the hoop. We want to know how far the ball travels horizontally by the time it hits the hoop, or the distance between you and the hoop. This is going to look like this:

x = v0x*t. So, (18.79 m/s)*(2.399 s) = 45.08 m.

d) Ignoring air-resistance, the x-component of the final velocity should remain the same, 18.79. The y-component of the velocity will change. Originally, it is 6.84. The equation to find the final velocity looks like this:

v = v0 + a*t.

So, filling in, v = 6.84 + (-9.8)*(2.399) = -16.6702 m/s.

e) To find the final velocity, we use the pythagorean theorem.

v = (vx^2 + vy^2) = (18.79^2 + -16.6702^2) = 25.12 m/s.

To find the direction of this magnitude relative to the positive x-axis, we want to find the angle in the right triangle. = arctan(-16.6702/18.79) = -41.5789 degrees. You can leave this as your answer, or you can subtract 360 by 41.5789 to get 318.42 degrees. I think they are both acceptable answers, but the positive answer is preferrable.

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