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 degrees above the horizontal and with a speed of v initial = 4.88 m/s. The ball is released 1.83 m above the floor. Ignore air resistance. The shot misses.

For the second throw, the ball goes through the center of the basket. For this throw, the player again shoots at 35 degreees 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?

Explanation / Answer

vx0 = v0cos(theta) = 4.00 m/s vy0 = v0sin(theta) = 2.80 m/s Maximum height occurs when vy = 0. vy = vy0 -g*t t = vy0/g = 2.8/9.8 = 0.29 s y = y0 + vy0*t -0.5*g*t^2 = 1.83 + 2.8*0.29 - 0.5*9.8*0.29^2 a)y = 2.23 m b) y = 0 = 1.83 + 2.8*t -0.5*9.81*t^2 t = 0.96 s x = vxo*t = 4*0.96 = 3.84 m c) This is slightly more complicated. Start with the x direction: x = vx0*t, t = 4.21/vx0 at this time, the basketball needs to be the height of the basket. Subsitute t from above into the y equation y = 3.05 = 1.83 + vy0*t - 0.5*g*t^2 3.05 = 1.83 + v0sin(theta)*4.21/v0*cos(theta) - 0.5*g*(4.21/v0cos(theta))^2 1.22 = 2.95 - 130/v0^2 v0 = 8.67 m/s d) do this the same as b, with different #s vx0 = 7.10 m/s vy0 = 4.97 m/s 0 = 4.97 - gt, t = 0.51s y = 1.83 + 4.97*0.51 - 0.5*9.81*0.51^2 y = 3.09 m x = 7.10*0.51 = 3.62 m distance away from the basket is: 4.21 - 3.62 = 0.59 m

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