Question

. The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 8.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

Answer 1

Answer:

$\theta = 67.22 degree$

Explanation:

Let say the ball is projected at an angle with horizontal

So here two components of the velocity of the ball is given as

$v_x = 8.15 cos\theta$

$v_y = 8.15 sin\theta$

now the displacement in x direction is given as

$x = v_x t$

$4.57 = (8.15 cos\theta)t$

in y direction it is given as

$y = y_o + v_y t - \frac{1}{2}gt^2$

$3.05 = 2.44 + (8.15 sin\theta) t - 4.9 t^2$

now from above two equations

$0.61 = 4.57 tan\theta - 4.9(\frac{4.57}{8.15 cos\theta})^2$

$0.61 = 4.57 tan\theta - 1.54(1 + tan^2\theta)$

$1.54 tan^2\theta - 4.57 tan\theta + 2.15 = 0$

$\theta = 67.22 degree$