A basketball player shoots the ball from 6 feet above the ground towards a baske

Question

A basketball player shoots the ball from 6 feet above the ground towards a basket that is 10 feet above the ground.

Assume that the player is 26 feet away horizontally (instead of the 15 feet assumed for the illustration)

(a) Suppose she shoots the ball at an angle of degrees above the horizontal (0 ) with an initial speed V. Give the x- and y-coordinates of the 2 position of the basketball at time t. Assume the x-coordinate of the basket is 0 and that the x-coordinate of the shooter is 26. Ignore air resistance (for now) and use an acceleration of gravity g of 32 ft/sec2.

(b) Using the parametric equations you obtained in part (a), experiment with different values for V and plotting the path of the ball to see how close the ball comes to the basket. Find some values of V and for which the shot goes in.

(c) Find the angle that minimizes the velocity needed for the ball to reach the basket. (This is a lengthy computation. First find an equation in V and that holds if the path of the ball passes through the point 26 feet from the shooter and 10 feet above the ground. Then minimize V.)

Explanation / Answer

Distance traveled in y-direction = 4 feet with initial speed vsin(a)

Distance traveled in x-direction = 26 feet with initial speed vcos(a)

Newton's Second law :S = ut +1/2gt2

For vertical direction as gravity will act agains the motion: 4 = vsin(a)t - 1/2(32)t2 --------------------------------(1)

For horizontal direction with no resistance 26 = vcos(a)t -------------------------------------------------(2)

Using equation 1 and 2

4 = vsin(a)* 26/vcos(a) -16(26/vcos(a))2

4 =26tan(a) - 16*26*26 /v2 cos2(a)

Equation between angle (a) and velocity v : 4 =26tan(a) - 16*26*26 /v2 cos2(a)

26tan(a) -4 = 10816/v2cos2(a)

v2 = 10816/(26tan(a) -4)cos2(a)

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