help with physics please?

Question

On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5 times as far as he would have on earth, given the same velocities on both planets. The ball is launched at a speed of 56 m/s at an angle of 38° above the horizontal. When the ball lands, it is at the same level as the tee.

(a) On the distant planet, what is the maximum height of the ball?

(b) On the distant planet, what is the range of the ball?

Answer 1

OK. First break up the balls velocity vector into seperate horizontal and vertical components
vx = 56*cos(38) = 44.13
vy = 56*sin(38) = 34.48
Then calculate the maximum height of the ball. Remember
v² = vy² - 2gy (where x is velocity, g is gravitational acceleration, and vy is initial velocity) At the highest point, v = 0 so
34.48² = 2gy But you don't know g for the 'faraway planet'. But you *do* know is that the ball is driven 3.5 times as far. So...... Let's see how it would play out on Earth. ON Earth, the ball would go
34.48²/(2*9.8) = 60.66 meters into the air and, since
s = gt²/2 it takes 3.52 seconds to get to it's maximum altitude and 3.52 seconds to fall back. During this 7.04 seconds, the ball (traveling at 34.48 m/s) travels
34.48*7.04 = 242.7 meters.

Now..... You know that on the 'faraway planet' it travels
3.5*242.7 = 849.5 meters. But its horizontal velocity was the same. Therefore, it was in the air
3.5*7.04 = 24.64 seconds and it took the same amount of time to get to the top of it's arc as it took to fall back.

Using this information, you can now go back and do a wee bit of algebra with the equation(s) used to get the height and figure that one out by yourslef ☺

Hope that helped.


Doug

Answer 2

This is hard!