A golf ball is hit with an initial velocity of 50 m/s at an angle of 39 degrees above the horizontal. How far does it travel?(horizontal distance)(g = 9.8 m/s2).
I made a right triangle. Found that Vi(x)=38.857, Vi(y)=312.466, both Vf's were 0, although Vf(y) is taking only have the projectile into effect (peak of flight).
Found the time taken by 0=31.466+-9.81t (Vf=Vi+at). T=3.207s.
Doubling the time to take the full flight into effect I did d=(38.857/2) * 6.415.
D=124.634.
Did I do it right? My answer is incorrect. Any input is greatly appreciated.
50 sin 39 = 31.5 m/s up
at peak v = 0
0 = 31.5 - 9.81 t
t = 3.207 check with you at peak
2 t = 6.415 check
horizontal speed constant = 50 cos 39 = 38.86 m/s
speed * time = 38.86*6.415 = 249.3
Hey, it is in the air the whole time, not just half the time. It is still going forward while it is falling.