Question

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?

Answer 1

Yo = 26 m/s.=Ver. component of initial velocity.

Y = Yo + gt.
Tr = (Y-Yo)/g = (0-26/-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.

Answer 2

4.5s