Question

0 12" 12' [8] A fireworks shel I is launched vertically from point A with speed sufficient to reach a maximum altitude of 500 ft. A steady horizontal wind causes a horizontal acceleration of 0.5 ft/s 2 without affecting the vertical motion, what angle α compensates for the wind such th at the shell peaks directly over the launch point A? maximum height h achievable when launched at this angle? What is the Wind- MacBook Pro

Answer 1

Solution,

Using 3rd equation of motion,

V^2 = u^2 + 2ah

0 = u^2 + 2 (-9.8) 500

u = 99 ft/s

For compensation,

g sin(alpha) = 0.5

Alpha = 2.92 degree

Maximum height,

0 = 99^2 - 2 x 9.8 x cos(2.92) x h

Height, h = 500.7 ft

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