Answer to Question #233289 in Mechanics | Relativity for Sum

Question

On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 5.00 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 42.5 m/s at an angle of 26 ° above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet, what are (a) the maximum height and (b) the range of the ball?

Answer 1

As the ball launched and landed at the same vertical level which means that the vertical displacement is zero Y = 0, by neglecting the air resistance the ball will experience an acceleration in the vertical direction which is the acceleration of the gravity, so the velocity in the horizontal direction is constant and its given with:

$v_{0x}=\frac{x}{t} \\ t = \frac{x}{v_{0x}} \; \;\;(1)$

in the vertical direction, the following equation of motion describes the motion in this direction:

$y=v_{0y}t + \frac{1}{2}at^2 \\ v_{0y}= -\frac{1}{2}at \;\;\; (2)$

substitute from (1) into (2) to get:

$v_{0y}= -\frac{1}{2}a(\frac{x}{v_{0x}})\;\;\;(3)$

the horizontal and the vertical components of the velocity are given by:

$v_{0x}=v_0cos(θ) \\ v_{0y}=v_0sin( θ)$

substitute into (1) to get:

$v_0sin( θ) = -\frac{1}{2}a(\frac{x}{v_0cos(θ)})$

rewrite for x,

$x=\frac{2v_0^2 sin(θ) cos(θ)}{a}$

this is how far the ball will travel in the horizontal axis, where A is the acceleration of the gravity, on earth $a_e=-9.8 \;m \cdot s^{-2}$ , so the distance on earth is:

$x_e=\frac{2v^2_{0,e} sin(θ_e) cos(θ_e)}{a_e}$

and the distance on the planet is (note a_e is the planet’s gravity acceleration):

$x_p=\frac{2v^2_{o,p} sin(θ_p) cos(θ_p)}{a_p}$

On the planet the ball 5.0 times as he would have on earth, given, the same initial velocities on both planets, which means $v_{0,e}=v_{0,p}=v_0$ and $θ_e= θ_p= θ$

$x_p=5.0x_e \\ \frac{v^2_0 sin(θ) cos(θ)}{a_p} = \frac{5 v^2_0 sin(θ) cos(θ)}{a_e} \\ \frac{1}{a_p}=\frac{5}{a_e} \\ a_p=\frac{1}{5}a_e$

if the acceleration of the gravity of earth has a value of $a_e=-9.8 \;m \cdot s^{-2}$ , the acceleration of gravity on the planet:

$a_p=-1.96 \;m \cdot s^{-2}$

(a) The maximum height occur when the vertical velocity is zero $v_y = 0$ , using this fact we can find the maximum height as:

$v^2_y=v^2_{0y}+2a_py_p \\ v^2_{0y}=-2a_p y_p \\ v^2_0sin^2( θ) = -2a_py_p \\ y_p= -\frac{v^2_0 sin^2(θ)}{2a_p}$

substitute with the given values to get:

$y_p = - \frac{(42.5 \;m/s)^2 sin^2(26)}{2(-1.96 \;m \cdot s^{-2}} \\ = 88.55 \;m$

(b) The range is given by the following equation, (we get it from part (a)):

$x_p=\frac{2v^2_0 sin(θ) cos(θ)}{a_p}$

substitute with the given values to get:

$x_p=\frac{2(42.5 \;m/s)^2 sin(26) cos(26)}{1.96 \; m/s^2} \\ = 726.25 \;m$

Answer: (a) $y_p= 88.55 \;m$ , (b) $x_p = 726.25 \;m$