Symon2-31

2-31 .\n\na) object starting at v0=0 and reaching max height\n\nv(0)=v0=0 t=0 t1 t2\n\nm dv\ndt = -mg - bv2\n\n-m dv = -mg - bv2 dt\n\ndv = -b dt\n\nm s + mg \n\ndv = -b dt\n\ng = {0 time for max}\n\n√(v2 )= ∫[0 to b]{dm/ms }\ndv/m \n=\n\ng = (8)..c\n\nv = sqrt(mg/b) tan (b(0 + t))\n\n∫dx = m/(g/b)[tan( v(0)t1 ) - tan( v(0)+t2)]\n\nx = m∫cos√( mg(b)) - a0\n=x = m/b ln{ ( 1 + b(v0 b ) ) -1 b/xs(x_)\n-x= (m/2b)ln[1 + b] -1/√( +2 tan (0 t1)\n2\n\nx = [n/b] ln[1 + 1] +n tan 1(\n\nx = m/b ln(1 (t1-0)) ln(0)) 2-31 .\n\nb): for t>t0, v=0\n\nmdv\ndt = -mg - b v2\n\nm dv = -mg dt - mbv2 dt\n\ndv = -b dt\n\nms v2\n\n\ndv = -b dt\n\ng = n = {b of zero}\n\ndx = m(b) dt + b \n√({n} = tanh\n} ... via nh x to slah.da2)\nin(\\\\\n=\n\ngx = tanh( ( −2) -c )\n\nx= -n m + ltanh( m + 1)e ( (t-t0)-tanh)\n\nx - xn = -n m / 2b ln cT ms{} glm csch\ntanh(t-t0)\n\ng = -m b tan h( /{}\n\n= m/b ln[1 + b( *i (88tn-g - t0)(/)\n\ng = m/b ln(1 + b )) - m(0) ln(cosh(v)\n\nx = m/b ln( 1 + b t0 ) - m/b ln(cosh(m(t-t0)))\n}