For how much more time was the ball in flight??

Question

On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 30.0 m/s at an angle 60.0 degrees above the horizontal.


How much farther did the ball travel on the moon than it would have on earth?


For how much more time was the ball in flight?

Answer 1

No, I prefer not to use or memorize formules, it's better to do it like this, so you will understand it better :

In the moon :

We have the velocity : 30 m/s and making and angle with the horizontal of 60º.

Then, you can find both components : Vx and Vy :

Vx = 30*cos(60) = 15 m/s

Vy = 30*sin(60) = 15*sqrt(3) m/s

In the moon = g = 9.8 / 6 = 1.63 m/s^2

Now, let's find the flying time, it's not that difficult, just find the time the ball takes to reach the maximum altitude :

Just using the vertical component of the velocity

Maximum altitude, Vf = 0 >>> final speed

0 = 15 - 1.63*t

t = 9.2 (s)

The total flying time = 2t = 18.4 s

then, the ball traveled : X = Vx*18.4 = 15*sqrt(3)*18.4 = 478 m

In the Earth : g = 9.8 m/s^2

So using the same :

0 = 15 - 9.8t

t = 1.53 s

total flying time = 3.06 s

then : the ball would have traveled : X = 15*sqrt(3)*3.06 = 80 m

Answers : in the moon : X = 478

in the Earth = 80

the ball traveled 398 more meters than in the earth.

in the moon the ball took : 9.2 s

in the earth the ball took : 1.53 s

The ball took 7.67 more seconds in the moon than in Earth



That's it

I think that's the best way to do it

Answer 2

g = 9.8/6 = 1.63
setting y = 0 = 30sin60 - 1/2(1.63)t^2
t^2 = 60sin60/1.63
t = 5.64 sec total time on the moon