A basketball player has free throws from the foul line. The center of the basket

Question

A basketball player has free throws from the foul line. The center of the basket is a horizontal distance of 4.21 m from the foul line and is a height of 3.05 m above the floor. For the first free throw he shoots the ball at an angle of 35 degreesabove the horizontal and with a speed of v0 = 4.88 m/s. The ball is released 1.83 m above the floor. Ignore air resistance. The shot misses.

(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?

For the second throw, the ball goes through the center of the basket. For this throw, the player again shoots at 35o above the horizontal and released it 1.83 m above the floor.

(c) What initial speed does the player give the ball on this second attempt?

(d) For this second throw, what is the maximum height of the ball?

(e) At the maximum height, how far is the ball horizontally from the basket? (10 marks)

Explanation / Answer

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)

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