Question

a golfer tees off and hits a golf ball at a speed of 31 m/s and an angle of 35 degrees. how far did the ball travel before hitting the ground (answer) meters

Answer 1

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:
$x = x_o + v_o_xt + \frac{at^2}{2}$

Since there is no acceleration along x-direction; therefore,
$x = x_o + v_o_xt$

Since $v_o_x = v_ocos \alpha$ and $x_o$=0; therefore above equation becomes,

$x = v_ocos \alpha t$ --- (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,

=> $y = y_o + v_o_yt - \frac{gt^2}{2}$
=> $t = \frac{2v_o_y}{g}$

Since $v_o_y = sin \alpha$; therefore above equation becomes,
$t = \frac{2v_osin \alpha }{g}$

Put the value of t in equation (A):

(A) => $x = v_ocos \alpha \frac{2v_osin \alpha }{g}$

Where x = Range = R, and $2sin \alpha cos \alpha = sin(2 \alpha )$; therefore above equation becomes:

=> $R = (v_o)^2 *\frac{sin(2 \alpha )}{g}$

Now, as:
$v_o = 31 m/s$

and $\alpha = 35$°
and g = 9.8 m/(s^2)

Hence,
$R = (31)^2 *\frac{sin(2 *35 )}{9.8}$

Ans: R = 92.15 meters.

-i