Problem 3 An NBA player throws a basketball at a horizontal distance of 10.0 m f

Question

Problem 3 An NBA player throws a basketball at a horizontal distance of 10.0 m from the center of the basket. The basket is at a height of 3.05 m above the floor as shown below. He shoots the ball at an angle 80 -40.0° with respect to the horizontal direction and releases it at the height of 2.00 m above the floor. (a) Please indicate clearly in words on the following figure for the origin of the x and y coordinates that you will set to solve this problem, (b) What i basket without the ball touching the rim (a swish!!)? (c) What is the maximum height with resp (d) How far horizontally is the ball from the basket when the ball is at the maximum height? (e) Find the time of flight of the ball from the point it leaves the hand of the player to it reaches itial speed vo must the player throw the ball so that it goes through the center of the ective to the floor reached by the ball? basket. 2.00 n 11 m

Explanation / Answer

(a) Take the origin as the foot of the NBA player and x - axis as the horizontal distance between the player and basket. The y -axis is along the vertical height of the player.

(b) Suppose 'v' is the initial velocity of the ball.

So, horizontal component of velocity, vx = v*cos40 = 0.77*v

Vertical component of velocity, vy = v*sin40 = 0.64*v

Horizontal distance between the player and the basket, R = 10 m

So, time taken by the ball to cover this distance, t = R / vx = 10 / 0.77v = 13.0 / v

In this time t, the ball has to cover a vertical distance of 3.05 - 2.00 = 1.05 m.

use the expresstion -

h = v*t + (1/2)*g*t^2

put the values -

1.05 = 0.64v*(13/v) + 0.5*(-9.8)*(13/v)^2

=> 1.05 = 8.32 - 828.1 / v^2

=> 828.1 / v^2 = 8.32 - 1.05 = 7.27

=> v^2 = 828.1 / 7.27 = 113.9

=> v = 10.67 m/s

Hence the initial speed of the ball should be 10.67 m/s

(c) Maximum height of the ball above the height of the player = vy^2 / (2*g) = (0.64*10.67)^2 / (2*9.8)

= 2.38 m

Therefore, maximum height of the ball with respect to the floor = 2.00 + 2.38 = 4.38 m

(d) Time of flight of the ball = t = R/ vx = 13.0 / v = 13.0 / 10.67 = 1.22 sec.

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