Question

A golf ball with an initial angle of 34 degrees lands exactly 240 m down the range on a level course.

a.) Neglecting air friction, what initial speed would achieve this result?

I was looking up solutions online and I stumbled across this explanation:
Let the initial speed be v
Initial vertical speed is v * sin(34)
Time of the flight of the ball will be [2v * sin(34)]/g
Initial horizontal speed is v * cos(34)
Then, they did the following:
The range = [2v * sin(34)]/g * (v * cos(34)
= v^2 sin * 2 * 34 / g
= v^2 sin(68)/g

This is the only part I need help with.

Answer 1

My question is how they got two things. 1) The formula for the time of the flight of the ball. Second. When they multiplied the the range function, they somehow eliminated the cos and were left with only a sin? Any help is appreciated!!!

Answer 2

Time of flight is "governed" by the length of time the ball is IN THE AIR. What goes up, very often comes down. That takes time. So, working out the time to rise to the full height, and then drop to the ground gives you that time.
THEN how FAR will the thing travel IN THAT TIME gives you the "range" of the projectile.
I tend to do these problems using KINEMATIC EQUATIONS, and taking advantage of the fact that the VERTICAL movement is AT RIGHT ANGLES to the HORIZONTAL movement - the two are INDEPENDENT.
From a glance at what you've written, you've resolved vertically an horizontally using sines and cosines. That's similar to what I've just written, I hope. A sine function and a cosine function are, I think, 90 degrees out of phase, or at right angles to each other as takes the mathies' fancy to bamboozle.
I think this is right.
bon chance et bon voyage
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