A particle is projected vertically upward from the top of a tower of height H with adjustable speed u. Consider its motion from projection until it strikes the ground. The value of u for which its average speed over the entire motion is minimum is:
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Question

A particle is projected vertically upward from the top of a tower of height H with adjustable speed u. Consider its motion from projection until it strikes the ground. The value of u for which its average speed over the entire motion is minimum is:

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G1 Physics · Accepted Answer

The particle first rises by h = u²/(2g). Total distance until impact is D = h + (H+h) = H + u²/g. Time to hit the ground is T = [u + √(u²+2gH)]/g. Hence average speed is average speed = (gH+u²)/[u+√(u²+2gH)]. Let r = u/√(gH). Apart from the constant factor √(gH), minimize f(r) = (1+r²)/[r+√(r²+2)]. Differentiating and setting df/dr=0 gives −r² + 2r√(r²+2) − 1 = 0. Squaring the equivalent relation 2r√(r²+2)=r²+1 gives 3r⁴ + 6r² − 1 = 0. Let z=r². Then 3z²+6z−1=0. The positive root is z = (2√3−3)…