Question.
A particle accelerates with acceleration of magnitude (a) for T seconds at an angle w away from the vertical. The particle then coasts through ballistically in an arc under the influence of gravity, g, through a maximum height of H above the ground, and then falls to the ground. Find (a) as a function of H, T, w, and g.
Answer
Distance on straight line travelled,
OA = (1/2) aT^2
=> Height reached = OA cosw = (1/2) aT^2 cosw
=> ramaining height to be reached
s = H - (1/2) aT^2 cosw
Velocity at A = aT
Vertical component of velocity at A
= aTcosw
Using v^2 = u^2 - 2gs
=> 0 = (aTcosw)^2 - 2g [H - (1/2) aT^2 cosw]
=> (Tcosw)^2 a^2 + (gT^2 cosw) a - 2gH = 0
=> a^2 + (g/cosw) a - 2gH / (T^2 cos^2 w) = 0
=> a = (1/2) [ - g/cosw + √[g^2 / cos^2 w + 8gH/ (T^2 cos^2 w)] ]
=> a = [g/(2cosw)] [ -1 + √(1 + 8H/gT^2)].
A particle accelerates with acceleration of magnitude (a) for T seconds at an angle w away from the vertical. The particle then coasts through ballistically in an arc under the influence of gravity, g, through a maximum height of H above the ground, and then falls to the ground. Find (a) as a function of H, T, w, and g.
Answer
Distance on straight line travelled,
OA = (1/2) aT^2
=> Height reached = OA cosw = (1/2) aT^2 cosw
=> ramaining height to be reached
s = H - (1/2) aT^2 cosw
Velocity at A = aT
Vertical component of velocity at A
= aTcosw
Using v^2 = u^2 - 2gs
=> 0 = (aTcosw)^2 - 2g [H - (1/2) aT^2 cosw]
=> (Tcosw)^2 a^2 + (gT^2 cosw) a - 2gH = 0
=> a^2 + (g/cosw) a - 2gH / (T^2 cos^2 w) = 0
=> a = (1/2) [ - g/cosw + √[g^2 / cos^2 w + 8gH/ (T^2 cos^2 w)] ]
=> a = [g/(2cosw)] [ -1 + √(1 + 8H/gT^2)].
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