A small cannonball with mass 4 kilograms is shot vertically upward with an initial velocity of 110 meters per second. If the air resistance is assumed to be directly proportional to the speed of the cannonball, a differential equation modeling the projectile velocity is m dv dt mg kv. 9.8 meters/second². Solve the differential equation for the velocity v(t). Don't forget to include the initial condition. v(t) = = Assume that k = 0.0025, and use g =

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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A small cannonball with mass 4 kilograms is shot vertically upward with an initial velocity of
110 meters per second. If the air resistance is assumed to be directly proportional to the
speed of the cannonball, a differential equation modeling the projectile velocity is
dv
m. = mg - kv.
dt
9.8 meters/second².
Solve the differential equation for the velocity v(t). Don't forget to include the initial
condition.
Assume that k
v(t)
=
=
=
0.0025, and use g
-
Integrate the velocity to obtain the height h(t) as a function of time. Assume the cannonball
is launched from ground level at t = = 0.
h(t):
Find the maximum height reached by the cannonball.
Max height =
Transcribed Image Text:A small cannonball with mass 4 kilograms is shot vertically upward with an initial velocity of 110 meters per second. If the air resistance is assumed to be directly proportional to the speed of the cannonball, a differential equation modeling the projectile velocity is dv m. = mg - kv. dt 9.8 meters/second². Solve the differential equation for the velocity v(t). Don't forget to include the initial condition. Assume that k v(t) = = = 0.0025, and use g - Integrate the velocity to obtain the height h(t) as a function of time. Assume the cannonball is launched from ground level at t = = 0. h(t): Find the maximum height reached by the cannonball. Max height =
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