If one ignores air resistance, then
horizontal distance= V*cosTheta * t
but t is dependent on time in the air.
vertical distance= VsinTheta*t-1/2 g t^2=0
or t= 2VsinTheta/g
putting that into horizontal distance
distance=2V^2cosThetasinTheta/g
but cosThetasinTheta=1/2 sin(2theta)
d=2V/g *sin(2theta)
to find max distance, we take the first derivative and set to zero
d/dtheta (2V/g*cos(2theta)*2)=0
or 2theta=90 degrees, or theta=45 deg for max. For min distance, theta=0
A golf ball is given an initial speed of 20 m/s and returns to ground level. Which launch angle above level ground results in the ball traveling the greatest horizontal distance? Neglect friction.
1. 60 degrees
2. 45
3. 30
4. 15
The answer is 45 but I was wondering how to get there. The process.
Thanks.
1 answer
2 answers