# physics

posted by Kevin

A projectile is fired at 30.0° above the horizontal. Its initial speed is equal to 172.5 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored. What is the maximum height reached by the projectile? At what time after being fired does the projectile reach this maximum height?

1. Kevin

So in projectile problems horizontal a=0 so the velocity is constant. So finding x and y components for initial velocity
X: 172.5m/s*cos30=149.4m/s
Y: 172.5m/s*sin30=86.25m/s

The max height is at the apex of the projectile, so use vertical velocity in
Vfy =Viy+at---> -vi/a=t
-86.25m/s)/-9.8m/s^2=8.80 sec

Y=yo+vot+1/2at^2
Y= 87.25m/s*8.8s +.5*9.8m/s^2*8.8s^2
Y= 759m+379.456m
Y= 1138m high

Hope this helps!
[vfy=0 at apex]

2. Henry

Vo = 172.5m/s[30o].
Xo = 172.5*Cos30 = 149.4 m/s.
Yo = 172.5*sin30 = 86.25 m/s.

a. Yf^2 = Yo^2 + 2g*h.
h = -(Yo)^2/2g = -(86.25)^2/19.6 = 379.5 m.

b. Y = Yo + g*Tr = 0.
Tr = -Yo/g = -86.25/-9.8 = 8.80 s. = Rise time.

NOTE: In your Eq, use (-)9.8 for g.

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