Question

Chris threw a basketball a distance of 27.5 m to score and win his

high school basketball game. If the shot was made at a 50.0° angle
above the horizontal, what was the initial speed of the ball?

Answer 1

v₀ = 16.55 m/s

Step-by-step explanation:

This motion of the ball can be modeled as a projectile motion with following data:

R = Range of Projectile = 27.5 m

θ = Launch Angle = 50°

g = acceleration due to gravity = 9.81 m/s²

v₀ = Initial Speed of Ball = ?

Therefore, using formula for range of projectile, we have:

`R = (v_(0)^2\ Sin2\theta)/(g)\\\\v_(0)^2 = (Rg)/(Sin2\theta)\\\\v_(0)^2 = ((27.5\ m)(9.81\ m/s^2))/(Sin100^o)\\\\v_(0) = √(273.93\ m^2/s^2)`

v₀ = 16.55 m/s