A projectile is fired from the origin (at y = 0 m) as shown in the diagram.

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A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity components are  V 0x = 310 m/s and V 0y = 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. When the projectile reaches point Q, what is the vertical component of its velocity?

May 18th, 2015

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

V = Vo + gt.
Tr = (V-Vo)/g = (0-26)/-9.8 = 2.65 s. = Rise time.

Tf1 = Tr = 2.65 s. = Fall time to launching point.

h = Vo*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.

May 18th, 2015

that answer its not in my  answer choices

a.  - 40 m/s

b.  -26 m/s

c.  -87 m/s

d.  -33 m/s

3.  - 70 m/s


May 18th, 2015

one moment

May 18th, 2015

ok.

May 18th, 2015

(V y)^2 = (V 0y)^2 + 2g*h

(V y)^2 = 26^2 + 2*9.8*20=676+392=1068

V y = sqrt(1068)=32.68 = 33

V 0y  is up and V y is down, so the sign is different.

so answer is d) -33m/s

May 18th, 2015

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