Unformatted text preview: 54. (a) Since Sample Problem 159 deals with a similar situation, we use the ﬁnal equation (labeled “Answer”) from it:
v = 2gh =⇒ v = vo for the projectile motion.
The stream of water emerges horizontally (θo = 0◦ in the notation of Chapter 4), and setting
y − yo = −(H − h) in Eq. 422, we obtain the “timeofﬂight”
t= −2(H − h)
=
−g 2
(H − h) .
g Using this in Eq. 421, where xo = 0 by choice of coordinate origin, we ﬁnd
x = vo t = 2gh 2
(H − h) = 2 h(H − h) .
g (b) The result of part (a) (which, when squared, reads x2 = 4h(H − h)) is a quadratic equation for h
once x and H are speciﬁed. Two solutions for h are therefore mathematically possible, but are they
both physically possible? For instance, are both solutions positive and less than H ? We employ
the quadratic formula:
√
x2
H ± H 2 − x2
2
h − Hh +
= 0 =⇒ h =
4
2
which permits us to see that both roots are physically possible, so long as x < H . Labeling the
larger root h1 (where the plus sign is chosen) and the smaller root as h2 (where the minus sign is
chosen), then we note that their sum is simply
√
√
H + H 2 − x2
H − H 2 − x2
h1 + h2 =
+
=H .
2
2
Thus, one root is related to the other (generically labeled h and h) by h = H − h.
(c) We wish to maximize the function f = x2 = 4h(H − h). We diﬀerentiate with respect to h and set
equal to zero to obtain
df
H
= 4H − 8h = 0 =⇒ h =
dh
2
as the depth from which an emerging stream of water will travel the maximum horizontal distance. ...
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 Fall '08
 SPRUNGER
 Physics, Projectile Motion

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