Ans: R = Ball Travelled = 92.15 meters.
Explanation:
First we need to derive that formula for the "range" in order to know how far the ball traveled before hitting the ground.
Along x-axis, equation would be:
[tex]x = x_o + v_o_xt + \frac{at^2}{2} [/tex]
Since there is no acceleration along x-direction; therefore,
[tex]x = x_o + v_o_xt[/tex]
Since [tex]v_o_x = v_ocos \alpha [/tex] and [tex]x_o[/tex]=0; therefore above equation becomes,
[tex]x = v_ocos \alpha t[/tex] --- (A)
Now we need to find "t", and the time is not given. In order to do so, we shall use the y-direction motion equation. Before hitting the ground y ≈ 0 and a = -g; therefore,
=> [tex]y = y_o + v_o_yt - \frac{gt^2}{2} [/tex]
=> [tex]t = \frac{2v_o_y}{g} [/tex]
Since [tex]v_o_y = sin \alpha [/tex]; therefore above equation becomes,
[tex]t = \frac{2v_osin \alpha }{g} [/tex]
Put the value of t in equation (A):
(A) => [tex]x = v_ocos \alpha \frac{2v_osin \alpha }{g} [/tex]
Where x = Range = R, and [tex]2sin \alpha cos \alpha = sin(2 \alpha )[/tex]; therefore above equation becomes:
=> [tex]R = (v_o)^2 *\frac{sin(2 \alpha )}{g} [/tex]
Now, as:
[tex]v_o = 31 m/s[/tex]
and [tex] \alpha = 35[/tex]°
and g = 9.8 m/(s^2)
Hence,
[tex]R = (31)^2 *\frac{sin(2 *35 )}{9.8}[/tex]
Ans: R = 92.15 meters.
-i
