Suppose a small cannon weighing 16 pounds is shot vertically upward with an init

Question

Suppose a small cannon weighing 16 pounds is shot vertically upward with an initial velocity v_0 = 300 ft/sec. the answer to the questions "How high does the cannonball go?" depends on whether we take air resistance into account. Take the positive direction to be upward. Complete the following for the three cases below: Write a model for the velocity V (t) of the object and solve for the velocity of the cannonball. Use this result to determine the height s (t) of the cannonball measure from ground level. Find the maximum height of the cannonball. Solve the problem above three times, for each of the cases below. Assume air resistance is proportional to instantaneous velocity with constant of proportionality k = 0.0025 Assume air resistance is proportional to the square of the instantaneous velocity with constant of proportionality k = 0.0003

Explanation / Answer

given,

v0=300ft/s             M=16 pounds

for case 1,

a)

in this case we can diractly apply eqn. of motion

v(t)=v0+at   (Eqn 1)

in this case the only force acting is gravity so,

a=g=-9.8m/s2=-32.15ft/s2

so,

v(t)=300-32.15t     (in ft/s)   (Eqn. 1A)

b)

according to equation of motion,

v(t)2=v02+2as(t)    (Eqn.2)

if we apply Eqn. 1--->Eqn. 2 we will get

s(t)=v0t+0.5at2   (Eqn. 3)

after substituting the value of a,v0

s(t)=300t-16.075t2   (in ft)    (Eqn. 3A)

c)

at maximum hight v(t)=0

so from Eqn.1A

0=300-32.15t

t=9.33 s

after substituting to Eqn. 3A

s(9.33)=300*9.33-16.075*9.332

s(9.33)=1399.69 ft

...............................................................................................

for case 2,

a)

the net force on the ball

F(t)=Mg-kv(t) (Eqn. 4)

since the force is varing with time acceleration also be a function of time so,

Ma(t)=Mg-kv(t)

a(t)=g-(k/M)v(t)

by applying to Eqn. 1

v(t)=v0+(g-(k/M)v(t)) t

by simplification,

v(t)=(v0+gt)/(1+(k/M) t)   (Eqn. 5)

by giving the values k=.0025 , M=16 pounds ,g=-32.15ft/s2 and v0=300ft/s

v(t)=(300-32.15t)/(1+1.5625*10-4*t)

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