Respuesta :
Answer:
[tex] 32 [/tex] m/s
Explanation:
[tex]\theta[/tex] = angle of launch of the ball = 25 deg
[tex]v_{o}[/tex] = speed of launch of the golf ball = ?
Consider the motion along vertical direction
[tex]v_{oy}[/tex] = initial velocity along y-direction = [tex]v_{o} Sin\theta[/tex] = [tex]v_{o} Sin25[/tex]
[tex]t[/tex] = time of travel in air = ?
[tex]Y[/tex] = vertical displacement = 0 m
[tex]a[/tex] = acceleration due to gravity = 9.8 m/s²
Using the equation
[tex]Y = v_{oy} t - (0.5) a t^{2}[/tex]
[tex]0 = v_{o} Sin\theta t - (0.5) a t^{2}[/tex]
[tex]t = \frac{2v_{o} Sin\theta }{a}[/tex] eq-1
Consider the motion along horizontal direction
[tex]v_{ox}[/tex] = initial velocity along x-direction = [tex]v_{o} Cos\theta[/tex] = [tex]v_{o} Cos25[/tex]
[tex]t[/tex] = time of travel in air = ?
[tex]X[/tex] = horizontal displacement = 80.0438 m
Using the equation
[tex]X = (v_{o} Cos\theta) t[/tex]
[tex]80.0438 = (v_{o} Cos25) t[/tex]
using eq-1
[tex]80.0438 = (v_{o} Cos25) \left ( \frac{2v_{o} Sin25 }{g} \right )[/tex]
[tex]80.0438 = (v_{o} Cos25) \left ( \frac{2v_{o} Sin25 }{9.8} \right )[/tex]
[tex]v_{o} = 32 [/tex] m/s
