Let mass, height, velocity, time are respectively represented by m, v, s , t.
So mg = 98N where g = 9.8 m/s²
So m = 98/9.8 = 10 kg
Initial velocity v0 = 82 m/s
(A)
A model of the state of the cannon ball is given by
mdtdv = - mg - kv
=> mdtdv = -k(v + kmg )
=> (v+kmg)dv = −mkdt
Integrating,
∫ (v+kmg)dv = ∫ −mkdt
=> ln(v+kmg) = −mkt + C
By initial condition, when t=0, v= 82
So ln(82+kmg) = C
So ln(v+kmg) = −mkt + ln(82+kmg)
=> ln(v+kmg) - ln(82+kmg) = −mkt
=> ln (82+kmg)(v+kmg) = −mkt
=> (82+kmg)(v+kmg) = e−mkt
=> (v+kmg) = (82+kmg) e−mkt
=> v=−kmg + (82+kmg) e−mkt
Putting the values of m,k,g we get
kmg=0.00259.8∗10=39200
mk=0.00025
So expression for velocity is
v(t) = -39200+39282e−0.00025t
(B)
From part - A we can write
dtds = -39200+39282e−0.00025t
=> ds = (-39200 + 39282e−0.00025t )dt
∫ ds = ∫(-39200 + 39282e−0.00025t )dt
=> s = -39200t - 0.0002539282 e−0.00025t + C1
By initial condition, s = 0 when t=0
C1 = 0.0002539282=157128000
So s = 157128000 - 39200t - 157128000e−0.00025t
So height of the cannon ball is
s(t)=157128000(1- e−0.00025t ) - 39200t
(C)
For maximum height ,dtds=v=0
=> -39200 + 39282e−0.00025t=0
=> e−0.00025t=3928239200
=> -0.00025t = ln(3928239200 )
=> -0.00025t = - 0.00208965189
=> t = 0.000250.00208965189 = 8.35860756
Putting this value of t in s(t) we get maximum height.
So maximum height is
157128000(1 - e−0.00025∗8.35860756 ) - 39200*8.35860756
= 328000-327657.416
= 342.584
So maximum height is 342.584 m
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