A fireworks shell is launched vertically from point A with speed v_0 sufficient

Question

A fireworks shell is launched vertically from point A with speed v_0 sufficient to reach a maximum altitude of 500ft. A steady horizontal wind causes a constant horizontal acceleration of 0.5 f/sec^2, but does not affect the vertical motion. Assume g = 32.2 ft/s^2. Determine: (a) Figure 1: The deviation, delta, at the top of the trajectory caused by the wind (b) Figure 1: The vertical speed, v_0, required for this launch. (c) Figure 2: At what launch angle, alpha, is the shell to be launched at twice the initial speed v_0, so that it reaches its maximum height directly above the launch point, A? (That is, delta = 0)

Explanation / Answer

(A) at maximum altitude, vy = 0

Applying vf^2 - vi^2 = 2 a d in vertical,

0^2 - v0y^2 = 2 x -32.2 x 500

v0y = 179.44 ft/s

Applying vf = vi + a t

0 = 179.44 - 32.2t

t = 5.57 sec

deviation = 0 + (0.5 x 5.57^2 /2 )

= 7.76 ft .....Ans

(b) initial vertical speed = 179.44 ft/s

(c) initial speed = 358.88 ft/s

first lets calculate time taken to reach the maximum height,

0 = 358.88 cos(alpha) - 32.2 t

t = 11.15 cos(alpha)

in this time, displacement in horizontal direction will be zero.

d = v0 t + a t^2 /2

0 = 358.88 sin(alpha) t - 0.5 t^2 /2

t ( 717.76 sin(alpha) - 0.5 t) = 0

t = 717.76 sin(alpha) / 0.5

=> 11.15 cos(alpha) = 1435,52 sin(alpha)

tan(alpha) = 11.15 / 1435.52 = 0.445 deg ....Ans

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