Question

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]

Answer 1

First, we calculate the time for the projectile to stop and then we can get the maximum height:

$V_0=50\frac{m}{s} \\ V_{(flight)}=V_0-gt \implies t=V_0/g \\ H_{(flight)}=V_0t-\frac{1}{2}gt^2=t(V_0-\frac{1}{2}gt) \\ H_{(flight)}=(V_0/g)(V_0-\frac{1}{2}V_0) \\ \therefore H_{(flight)}=\cfrac{V^2_0}{2g}=\cfrac{(50\frac{m}{s})^2}{2(9.81\frac{m}{s^2})}=127.42\,m$

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):

$\sum F_{external}=ma=-F_D-mg=-(0.01\frac{kg}{m})V^2-(10\,kg)g \\ \implies a= \cfrac{dV}{dt}= \cfrac{dV}{dy}\cfrac{dy}{dt} \\\implies a=V\cfrac{dV}{dy}=-((0.001\,{m}^{-1})V^2+g) \\ \int_{V_0}^{V=0} \cfrac{V\,dV}{(0.001\,{m}^{-1})V^2+g}=- \int_{y=0}^{y=H_{(flight)}} dy \\ \bigg[ \frac{1}{(0.002\,{m}^{-1})} \ln{ ({(0.001\,{m}^{-1})V^2+g} ) } \bigg]_{V_0}^{V=0} =-H_{(flight)} \\ \text{ } \\ H_{(flight)}=-(500\,m) (\ln{ (g ) }-\ln{ ({0.001V_0^2+g} ) } ) \\ H_{(flight)}=(500\,m) \times \ln{ \cfrac{{0.001(50)^2+9.81}} {(9.81)} }=113.50\,m$

In conclusion, the maximum height of the projectile is (a) 127.42 m without air resistance and (b) 113.50 m when it is considered that F_D= 0.01V².

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.