Question

After being assaulted by flying cannonballs, the knights on the castle walls (12.0 m above the ground) respond by propelling flaming pitch balls at their assailants. One ball lands on the ground at a distance of 81.1 m from the castle walls. If it was launched at an angle of 53.0° above the horizontal, what was its initial speed?

Answer 1

28.6 m/s

Step-by-step explanation:

Using the equation for the range of a projectile,

R = U²sin2θ/g where U = initial speed of flaming pitch balls, θ = launch angle = 53° and g = acceleration due to gravity = 9.8 m/s²

Making U subject of the formula, we have

U = √(gR/sin2θ)

substituting the values of the variables into the equation given that R = 81.1 m, we have

U = √(9.8 m/s² × 81.1 m/sin2(53°))

U = √(794.78 m²/s²/sin106°)

U = √(794.78 m²/s²/0.9613)

U = √(826.78 m²/s²)

U = 28.75 m/s

U ≅ 28.6 m/s