physics
posted by Anonymous .
A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity components are V0x = 310 m/s and V0y = 26 m/s. The projectile reaches maximum height at point P, then it falls and strikes the ground at point Q, which is 20 m below the launch point. For how long is the projectile in the air?

Yo = 26 m/s.=Ver. component of initial velocity.
Y = Yo + gt.
Tr = (YYo)/g = (026/9.8 = 2.65 s. =
Rise time.
Tf1 = Tr = 2.65 s. = Fall time to launching point.
h = Yo*t + 0.5g*t^2 = 20 m.
26t + 4.9t^2 = 20
4.9t^2 + 26t  20 = 0
Use Quad. Formula:
Tf2 = 0.68 s. To fall 20 m.
T = Tr + Tf1 + Tf2=2.65 + 2.65 + 0.68 =
5.98 s. = Time in air. 
4.5s
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