I am trying to design a device that uses a spring to shot a projectile and I am

Question

I am trying to design a device that uses a spring to shot a projectile and I am having difficulty figuring out the angle and initial speed needed. I need it to go a horizontal distance of 30 ft with a velocity of 29 ft/s^2 at impact. The height difference between the initial and final locations is 32 inches (final location being higher). I know this is simple, I am simply decades out of practice.

I know the horizontal velocity is the same throughout as I am simplifying by not accounting for wind or drag. The horizontal acceleration is -32ft/s^s

X=Xo+Vxt Y=Yo+Vyt-.5gt^2 tan(theta)=Vy/Vx Vy=(Vy)o - gt

All the equations seem to have more than one unknown.

Explanation / Answer

horizontal distance d = 30 ft, final velocity v = 29 ft/s. h = 32 inches = 32/12 ft = 8/3 ft

find initial velocity v0 and angle

vertical displacement h = v0sin*t - gt2/2 (1)

horizontal displacement d = v0cos*t (2)

final velocity: vx = v0cos, vy = v0sin - gt, so v2 = vx2 + vy2 = (v0cos)2 + (v0sin - gt)2 = v02 - 2v0sin*gt + g2t2 = v02 - 2g(v0sint + gt2/2) = v02 - 2gh

v0 = (v2 + 2gh) = 31.8 ft/s

from (2), t = d/(v0cos)

then put it in (1), h = dtan - 0.5gd2/(v02cos2) = dtan - 0.5gd2sec2/v02 = dtan - 0.5gd2(1 + tan2)/v02

0.5gd2(1 + tan2) - dv02tan + hv02 = 0

0.5gd2tan2 - dv02tan + (hv02 + 0.5gd2) = 0

16*302tan2 - 30*31.82tan + (8/3*31.82 + 16*302) = 0

14400 tan2 - 30337 tan + 17097 = 0

no solution. It means impossible to shot it with such d, h and v

If 29 ft/s is the x component of final velocity, then v0x = vx = v0cos = 29 ft/s

t = d/vx = 1.03 s, from (1), v0sin = h/t + gt/2 = 19.1 ft/s

v0 = 34.7 ft/s, tan = 19.1/29, = 58.2o

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