Page 57 - How To Solve Word Problems In Calculus

(d) The maximum height is reached when the projectile’s velocity
is 0.
v(t) = 80 − 32t

0 = 80 − 32t
80 5
t = =
32 2
h(5/2) = 196 ft

The maximum height reached is 196 ft above the ground.
(e) The projectile returns to the ground when h(t) = 0.

2
h(t) = 80t − 16t + 96
2
0 = 80t − 16t + 96
2
0 = 5t − t + 6
2
0 =−(t − 5t − 6)
0 =−(t − 6)(t + 1)

t =−1 t = 6

We reject the first solution. The projectile hits the ground 6
seconds after it is thrown.

v(6) = 80 − 32 · 6 =−112 ft/sec

5. Let m(x) represent the mass of the rod from its end until point x.
3
Then m(x) = kx . Since the total mass is 500 slugs,
m(5) = 500
3
k · 5 = 500
k = 4

Therefore, m(x) = 4x 3
(a) m(0) = 0 and m(5) = 500

m(5) − m(0) 500 − 0
ρ av = = = 100 slugs/ft
5 − 0 5 − 0

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