A 10-kg projectile is fired vertically upward from the ground, with an initial velocity of 50 m/s, as shown below. Determine the maximum height to which it will travel if (a) atmospheric resistance is neglected; and (b) atmospheric resistance is measured as FD = (0.01v2 ) N, where v is the speed of the projectile at any instant, measured in m/s. [10]
First, we calculate the time for the projectile to stop and then we can get the maximum height:
"V_0=50\\frac{m}{s}\n\\\\ V_{(flight)}=V_0-gt \\implies t=V_0\/g\n\\\\ H_{(flight)}=V_0t-\\frac{1}{2}gt^2=t(V_0-\\frac{1}{2}gt)\n\\\\ H_{(flight)}=(V_0\/g)(V_0-\\frac{1}{2}V_0)\n\\\\ \\therefore H_{(flight)}=\\cfrac{V^2_0}{2g}=\\cfrac{(50\\frac{m}{s})^2}{2(9.81\\frac{m}{s^2})}=127.42\\,m"
For the second part, we have to use the sum of external forces equal to the mass times the acceleration of the object, then we have two external forces: gravity and air resistance, we use this and the following operations to find the relation between velocity and height (we also have to consider that both forces have a negative sign because they are both against the movement of the projectile):
"\\sum F_{external}=ma=-F_D-mg=-(0.01\\frac{kg}{m})V^2-(10\\,kg)g\n\\\\ \\implies a= \\cfrac{dV}{dt}= \\cfrac{dV}{dy}\\cfrac{dy}{dt}\n\\\\\\implies a=V\\cfrac{dV}{dy}=-((0.001\\,{m}^{-1})V^2+g)\n\\\\ \\int_{V_0}^{V=0} \\cfrac{V\\,dV}{(0.001\\,{m}^{-1})V^2+g}=- \\int_{y=0}^{y=H_{(flight)}} dy\n\\\\ \\bigg[ \\frac{1}{(0.002\\,{m}^{-1})} \\ln{ ({(0.001\\,{m}^{-1})V^2+g} ) } \\bigg]_{V_0}^{V=0} =-H_{(flight)}\n\\\\ \\text{ }\n\\\\ H_{(flight)}=-(500\\,m) (\\ln{ (g ) }-\\ln{ ({0.001V_0^2+g} ) } )\n\\\\ H_{(flight)}=(500\\,m) \\times \\ln{ \\cfrac{{0.001(50)^2+9.81}} {(9.81)} }=113.50\\,m"
Reference:
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