Homework Answers
given, L2 = 296 cm
launch angle theta = 30 deg
components of initial velocity
vx = 2.8 m/s
vy = 4.84 m/s
h1 = 78.4 cm
h2 = 42.5 cm
a. time of flight = t
then vx*t = L2
t = L2/vx = 296(10^-2)/2.8 = 1.05714 s
at this time t
vx = 2.8 m/s
vy = voy*t - 0.5gt^2
vy = 4.84*1.05714 - 4.9*1.05714^2
vy = -0.3593 m/s
b. magnitude of velocuty = sqroot(vx^2 + vy^2) = sqroot(2.8^2 +
0.3593^2) = 2.822 m/s
direction = arctan(vy/vx) = qrctan(0.3593/2.8) = 7.312
deg below horizontal
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