Max. height and range problem I can't crack

[DuffBeer]

19.[2pt]
On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 4.00 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 42.6m/s at an angle of 28.2o above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet, what is the maximum height of the ball?
Answer:

20.[2pt]
What is the range of the ball?
Answer:


its so hard to do without Earth gravity... i can't figure it out! :-( please help!

 

[Sirus]

You're right, it is hard to do without a gravitational constant. Can you find the gravitational constant from the info you have?

 

[Tide]

The range is: R = \frac {v_0^2 \sin {2\theta}}{2g} so it's easy to infer that g on the planet is one fourth that on Earth.

The maximum height is H = \frac {v_0^2 \sin ^2 \theta}{2g} and you should have no trouble doing the rest.

 

[DuffBeer]

The range is: R = \frac {v_0^2 \sin {2\theta}}{2g} so it's easy to infer that g on the planet is one fourth that on Earth.

The maximum height is H = \frac {v_0^2 \sin ^2 \theta}{2g} and you should have no trouble doing the rest.

the trig identity is sin2(thet) is 2sin(thet)cos(thet) right?

my ansewrs not coming out correct... i quadroople checked the algebra

 

[HallsofIvy]

It's difficult for us to comment on what you have done when you haven't shown us what you have done.

 

[DuffBeer]

It's difficult for us to comment on what you have done when you haven't shown us what you have done.

oh, sorry.

took R = v0^2 * sin2(theta) /2g

turned it into v0^2 * 2 * sin(theta) * cos(theta) / 2g (trig identity)

crossed out the 2s

plugged in 28 for theta, 42.6 for v0, and 2.45 (.25*9.8) for G. - solved.

got 307.042 meters - got the question wrong...

i guess it doesn't matter now because i missed the due date for the problem, but id still like to figure out how to do it for future reference.

im not exactly clear on why g is .25...

 

[Tide]

Start with
R = \frac {v_0^2}{2g} \sin 2\theta
and with identical initial speed and angles just set up the proportion
\frac {R_1}{R_2} = \frac {g_2}{g_1}
Your statement was that R_2 = 4 R_1 so g_2 = \frac {1}{4} g_1.

 

[DuffBeer]

Start with
R = \frac {v_0^2}{2g} \sin 2\theta
and with identical initial speed and angles just set up the proportion
\frac {R_1}{R_2} = \frac {g_2}{g_1}
Your statement was that R_2 = 4 R_1 so g_2 = \frac {1}{4} g_1.

ohhh... thanks for clearing that up!