Question

Part 4) Solve for the velocity that the projectile was going just before the collision with the crossbar (12 pts) 210,0 cm Base this on your prediction of L2 from part 2, not the "Best L2” Use the value of time, "t", that you found for the collision with the crossbar (when predicting L2 in part 2). Calculate the velocity components ve and vy at that time. Then find the magnitude and direction of the velocity vector from the components. (Report the direction in standard position based on the coordinate system in the drawings.) Velocity direction = Velocity magnitude (speed ) =-

Part 4. Please show all work

Answer 1

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