Question

A few seconds after a plane took off it was hit by a missile launched from a distance on the ground. The path of the plane is described by y=(0.0005 x²) m. The missile traveled along a parabolic path described by y2 = (160 x), where the coordinates are measured in meters. If the components of the missile velocity in the horizontal and vertical directions are constant at vr =120 m/s and v; = 100 m/s, respectively, and the plane was rising with a constant upward velocity Vy = 12 m/s, determine the magnitudes of the velocity and acceleration of the plane the moment it was hit by the missile. у y= (0.0005 x) y2 = (160 x) X

Answer 1

Answers - Thu missile traveled along Given that The path of the plane is described by Y= 10.0005+2) parabolic path described y²=(160x) horizontal and vertical directions are constant at Von = 120 m/s The - 10 mis Y² = 1600 Y=(0.0005 x2) upwond with ry= 12 m/s the plane was rising now the plane at time of Collision dyldt Ny = 12m/s 0.001 * vx =12 s where xix horizontal velocity of the plane at the instant of collision X* VX - 12000

for the missile Vx ir constant x = 12om's y constant 12ml 160* axldt =160 120 zy vy = 160 + vx 2y * 100 Y = 96m from the equastion of motion of two plane 26/0.0005 = x2 X 437.198 0460 so for the plane vy izm's 12000/7 12000/439.17804600 27.326127995 ms from the path of plane Vy = 0.001x* vx ay = o.ool ex*ax ax + vx', )

ay = 0.001 (437 178 * ax + 27.3761 2787812) now plane was riving up with constant velocity so ay=o tlence ax -27.3741278752/438-193 -171633/7190973 misls ax = the plane velocity Hence at the time of collinory ir va spot ( vx?2 + vy) - 29.898 74 565 m Direction is arctant (vy/vx) san ardant (12/27-28) = 2346 deg x-anis Counterclocwise above acceleration in a = 471611 misls C-ve direction indiecting the accelation x direction in ve