Projectile motion problem. I need the solution and the answer. I'm really having a hard time figuring it out.

Question

A basketball player is fouled and knocked to the floor during a lay-up attempt. The player is awarded two free throws. The center of the basket is a horizontal distance of 4.21 meters (13.8 ft.) from the foul line and is a height of 3.05 meters (10 ft.) above the floor. On the first free-throw attempt, he shoots the ball at an angle of 35 degrees above the horizontal and with a speed of 4.88 m/s OR (16 ft./s) (initial velocity). The ball is released 1.83 meters (6 ft.) above the floor. This shot misses badly. Ignore air resistance.
a.) What is the maximum height reached by the ball?
b.) At what distance along the floor from the free-throw line does the ball land?
c.) For the second throw, the ball goes through the center of the basket. For this second free-throw, the player again shoots the ball at 35 degrees above the horizontal and releases it 1.83 meters above the floor.
What initial speed does the player give the ball on this second attempt?

d.) For the second throw, what is the maximum height reached by the ball? At this point, how far horizontally is the ball from the basket?

Answer 1

Since this is such a classical question, I'm not going to actually solve all of it for you. But I will tell you *how* to solve it and leave the arithmetic to you.

First, it's usually a good idea to draw a picture. Get a few pieces of graph paper and draw the origin in the lower left corner.

Now go 'up' the y axis to the point that corresponds to 1.83 meters and make a dot. This dot is where the ball is released.

Now go to 4.21 meters to the 'right' and up 3.05 meters and make another dot. This is the center of the basket.

When the ball is released, it has an initial velocity of 4.88 m/s at an angle of 35 degrees above horizontal. Yhis means that its velocity can be broken down into two vextor components (vertical and horizontal or y and x) as follows:

Vx = 4.88 cos(35) = 3.997 m/s horizontally
Vy = 4.88 sin(35) = 2.799 m/s vertically

Since there is no force acting on the ball horizontally, it will travel horizontally at a constant 3.997 m/s

Vertically, it has an acceleration of -9.8 m/s/s (gravity) acting upon it.

Since you're working this kind of problem, I assume that you've already had the equations of motion (involving position, velocity, acceleration, and time) along a straight line. This problem is no different than those problems except that you now have two lines along which you must calculate. The one (and only) variable that 'ties' the two axis together is time.

So......

Remember that velocity under acceleration is given (s a function of time) by

v(t) = v0 + at^2

If the vertical velocity is 2.799 m/s and gravity is acting 'downwards' (negative) at -9.8 m/s/s then the velocity of the ball along the vertical axis (as a function of time) is

Vx(t) = 2.799 - 9.8 t^2

At the top of it's trajectory the balls intaneous velocity is zero and that happens at

0 = 2.799 - 9.8 t^2

or

t = sqrt (2.799/9.8) = .534

seconds after the ball is released.

How high is the ball? Remember that

v^2 = v0^2 + 2ay

and we're interested at v = 0 so

0 = 2.799^2 - 2*9.8*y

so that

y = (2.799^2)/(2*9.8) = .399

meters above the release point or

.399 + 1.83 = 2.229 meters above the floor.

In .534 seconds at a velocity of 3.997 m/s horizontally, it will travel

.534 * 3.997 = 2.134 meters horizontally (to the right)

This is the 'top' of the balls trajectory (2.134,2.229)

Now I'm gonna let you figure out the rest of it. You only get to learn this stuff once and I already had my turn a bit over 40 years ago


Doug

Answer 2

it is not hard. believe me. in basic terms calls for in basic terms slightly algebra. What you will finally end up with is an equation looking some thing like this: [((v cos (fifty 3))/4)(v sin (fifty 3))] -9.8[(v cos (fifty 3))/4]^2 = 15 clean up for v for the 1st question, 0.5 it and use your kinimatics equations to make certain the place he smacks into the cliff partitions in the 2d. relaxing question, yet took me longer to type out than it may desire to have taken you to clean up.