A projectile is fired at an upward angle of 60 degrees with a speed of 100m/s. It lands on a plateau 150m higher. What is the projectile's speed the moment before it strikes the plateau?
I'm thinking to just use the kinematic equation Vf^2= Vi^2 + 2aΔy. Am I correct?
2 answers
since the ball drop 100m above the ground,we calculate max height to get the vy at the point of landing:h=(100sin60)^2/20=375m so it travels(375m-100m) from the max height to the top of building vertically:v^2=2gh=2*10*275=5500 v=74.1m/s since the vx=100cos60=50m/s is constant,so it land with resultant velocity of:sqrt(74.1^2+50^2)=89.4m/s
Vo = 100m/s[60o].
Xo = 100*Cos60 = 50 m/s.
Yo = 100*sin60 = 86.6 m/s.
Y^2 = Yo^2 + 2g*h = 86.6^2 - 19.6*150 =
4560.
Y = 67.5 m/s.
Xo = 100*Cos60 = 50 m/s.
Yo = 100*sin60 = 86.6 m/s.
Y^2 = Yo^2 + 2g*h = 86.6^2 - 19.6*150 =
4560.
Y = 67.5 m/s.
1 answer