Respuesta :
Answer:
The initial speed of the golf ball was 25 m/s.
Explanation:
To find the initial speed of a projectile motion we must first find the motion equation corresponding to each axis:
For X axis we have an Uniform Rectilinear Motion, since no forces act along it:
[tex]x_{(t)}=v_{ix}.t+x_i[/tex], where [tex]v_{ix}[/tex] stands for the initial speed component on x axis, and [tex]x_i[/tex] is the initial position on x axis (zero, in this case).
For Y axis we have a Constant Acceleration Motion, since the gravitational force acts along it:
[tex]y_{(t)}=-\frac{1}{2} g.t^{2} +v_{iy}.t+y_i[/tex], where g is the acceleration of gravity, [tex]v_{iy}[/tex] stands for the initial speed component on y axis, and [tex]y_i[/tex] is the initial position on y axis (zero, in this case).
We can describe the initial speed's components as follow:
[tex]v_{ix}=v_i.cos(30^0)[/tex] and [tex]v_{iy}=v_i.sin(30^0)[/tex]
Keeping this in mind, we're just three steps away from the answer.
The first step is expressing t in terms of x:
[tex]t=\frac{x}{v_i.cos(30^0)}[/tex]
The second step is replacing this expression for t on the y axis equation:
[tex]y_{(x)}=-\frac{1}{2} .g.(\frac{x}{v_i.cos(30^0)} )^2+\frac{v_i}{v_i} \frac{sin(30^0)}{cos(30^0)} .x[/tex]
Finally, we use the fact that when x=55.2312 m the golf ball lands, it means y=0 m, and resolve the equation for [tex]v_i[/tex]
[tex]0m=-\frac{1}{2} .9,8\frac{m}{s^2}.(\frac{55,2312m}{v_i.cos(30^0)} )^2+\frac{sin(30^0)}{cos(30^0)} .55,2312m[/tex]
Obtaining
[tex]v_i=25\frac{m}{s}[/tex]
Answer:
25 m/s
Explanation:
Angle of projection, θ = 30°
Horizontal distance, r = 55.231 m
Let u be the velocity of projection.
Use the formula of horizontal range
[tex]R=\frac{u^{2}Sin2\theta }{g}[/tex]
By substituting the values, we get
[tex]55.231=\frac{u^{2}Sin60 }{9.8}[/tex]
[tex]55.231=\frac{u^{2}\times 0.866 }{9.8}[/tex]
u = 25 m/s
thus, the initial velocity of golf ball is 25 m/s.
