Abasketball player shoots a free throw 2.00 m above the free throw line with a velocity of 8.09 m/s at 65.0°. the basket is 3.05m off the ground. a. solve for the range of the ball. (ans: 4.57 m) b. solve for the time the ball is in the air. (ans: 1.34 s)

a) The range of the ball is 4.58 m.

b) The ball is 1.34 s in the air.

Explanation:

The position of the ball at time "t" is given by the position vector r:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α - 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = throwing angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

a ) The range of the ball is the x-component of the vector "r final" in the figure (please, see attached figure).

At the time the ball is in the basket, the vertical component of the position vector will be (3.05 m - 2.00 m) 1.05 m (placing the origin of the frame of reference at the throwing point). Then, using the equation of the vertical position, we can calculate the time the ball is in the air and, with that time, we can calculate the range:

y = y0 + v0 · t · sin α - 1/2 · g · t²

1.05 m = 0 m + 8.09 m/s · t · sin 65° - 1/2 · 9.8 m/s² · t²

0 = - 1/2 · 9.8 m/s² · t² + 8.09 m/s · t · sin 65° - 1.05 m

Solving the quadratic equation:

t = 0.16 s and t = 1.34 s

The ball is at two times at that height: a short while after throwing (0.16 s) and when it is in the basket at 1.34 s.

Now, we can calculate the range of the ball using the equation of the x-component of the position vector:

x = x0 + v0 · t · cos α

x = 0m + 8.09 m/s · 1.34 s · cos 65.0°

x = 4.58 m

The range of the ball is 4.58 m.

b) The total time the ball is in the air was already calculated above: 1.34 s.

the maxium height is 34m/s