Answer:
v₀ = 16.55 m/s
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:
[tex]R = \frac{v_{0}^2\ Sin2\theta}{g}\\\\v_{0}^2 = \frac{Rg}{Sin2\theta}\\\\v_{0}^2 = \frac{(27.5\ m)(9.81\ m/s^2)}{Sin100^o}\\\\v_{0} = \sqrt{273.93\ m^2/s^2}[/tex]
v₀ = 16.55 m/s
