time in air comes from
hf=hi-vvertical*t-1/2 g t^2
but hf=hi=zero so you can solve for time in air.
Vvertical=v*sinTheta
so t=2VsinTheta/g
horizonal distance:
distance=vhorizontal*timeinair
= v costheta*t
so you have a final distance which has a cosTheta*sinTheta
in it. do you remember the double angle formula
Sin(2Theta)=cosTheta*sinTheta
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?
b.) Using the speed determined in item a, find the maximum height reached by the ball.
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
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!!!
1 answer