posted by Tsion on .
A projectile is launched from ground level at 37.0 m/s at an angle of 30.6 ° above horizontal. Use the launch point as the origin of your coordinate system.
(a) How much time elapses before the projectile is at a point 8.9 m above the ground and heading downwards toward the ground?
(b) How far downrange (the horizontal distance from the origin) was the projectile when it reached the highest point in its flight?
My answer for part a) 3.29 s
How do I get part b?
Vo = 37m/s[30.6o]
Xo = 37*cos30.6 = 31.8 m/s.
Yo = 37*sin30.6 = 18.8 m/s.
a. Yo*t + 0.5g*t^2 = 8.9 m.
18.8t - 4.9t^2 = 8.9
-4.9t^2 + 18.8t - 8.9 = 0
t = 3.28 s. and heading downward.
(Used Quad. formula).
b. Y = Yo - g*Tr = 0 at max ht.
18.8 - 9.8*Tr = 0
9.8Tr = 18.8
Tr = 1.92 s = Rise time or time to reach max ht.
Hor. Distance=Xo * Tr=31.8m/s * 1.92s =