Answer to Question #91267 in Mechanics | Relativity for Umar

Question

An athlete threw a basketball a distance of 27.5m to score and win the game. If the shot was made at a 50.0 degree angle above the horizontal, what was the initial speed of the ball? (16.6m/s)

Answer 1

We can find the initial speed of the ball from the projectile range equation:

$R = \dfrac{v_0^2sin2\theta}{g},$

here, $R = 27.5 m$ is the range of the projectile, $v_0$ is the initial speed of the ball, $\theta = 50^{\circ}$ is the launch angle and $g = 9.8 m/s^2$ is the acceleration due to gravity.

Then, from this formula we can calculate the initial velocity of the ball:

$v = \sqrt{\dfrac{Rg}{sin2\theta}} = \sqrt{\dfrac{27.5 m \cdot 9.8\dfrac{m}{s^2}}{sin2 \cdot 50^{\circ}}} = 16.6 \dfrac{m}{s}.$

Answer:

$v_0 = 16.6 \dfrac{m}{s}.$