Question

On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.80 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 46.7 m/s at an angle of 34 ° 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

in the earth

initial speed v_o = 46.7 m/s

= 34^o

range on earth  R_e = v_o²*sin(2)/g_e

R = (46.7)²*sin(68)/9.8

on the planet

initial speed v_o = 46.7 m/s

angle = 34^o

range on planet R_p = v_o²*sin(2)/g_p

R_p = (46.7)²*sin(68)/g_p

given R_p = 3.80*R_e

(46.7)²*sin(68)/g_p = 3.80*(46.7)²*sin(68)/9.8

g_p = 2.57
gravitational acceleration on planet is 2.57 m/s² .

(a)

maximum height on planet H = v_o²*(sin)²/(2*g_p)

H = (46.7)²*(sin34)²/(2*2.57)

H = 132.67 m

(b)

range Rp = v_o²*sin(2)/g_p

R_p = (46.7)²*(sin68)²/(2.57)

R_p = 729.51 m