Okay so max height:
y=v0yt-(1/2)gt2=(26m/s)(5.3s/2)-(1/2)(9.8m/s)(5.3/2)2
=34.4 m
For final velocity:
Vfx=15m/s because ax=0
Vfy=v0y-gt=26-(9.8)(5.3)=-25.94
Vf=√(152+(-25.94)2)=29.93m/s
Vf=29.93m/s 60o at the horizontal
Have I done it right?
So far this is how I've attempted to solve the problems
Initial Velocity
Vox=Vicosθ=(30m/s)(cos60o)=15m/s
Voy=Visinθ=(30m/s)(sin60o)=26m/s
Time used for ball to reach the ground
ΔVy=Vy-Voy=Voyt-1/2(g)(t2)
ΔVy=0=(30m/s)(sin60)=1/2(9.8m/s2)
=5.3s
Range
Range=Vxt=(30m/s)(cos60o)(5.3s)
=79.5m...
If a projectile is thrown at an angle of 60o and it's initial speed is 30m/s, find the maximum height, time used for the ball to reach the ground, the horizontal range the ball reached from starting point, and the final velocity.
So can i use 30m/s as the initial velocity?