A body of mass 5 slugs is dropped from a height of 100 ft with zero velocity. Assuming no air resistance, find (a) an expression for the velocity of the body at any time t, (b) an expression for the position of the body at any time t, and (c) the time required to reach the ground.

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Accepted Answer

Choose the coordinate system as in Fig. 7-5. Then, since there is no air resistance, (7.5) applies: dv/dt = g. This differential equation is linear or, in differential form, separable; its solution is v = gt+c. When t=0, v = 0 (initially the body has zero velocity); hence 0 = g(0) + c, or c = 0. Thus, v = gt or, assuming g = 32 [latex]ft / sec ^{2}[/latex] ,

[latex]v=32 t[/latex] (1)

(b) Recall that velocity is the time rate of change of displacement, designated here by x. Hence, v = dx/dt, and (1) becomes dx/dt = 32t. This differential equation is also both linear and separable; its solution is

[latex]x=16 t^{2}+c_{1}[/latex] (2)

But at t = 0, x = 0 (see Fig. 7-5). Thus, [latex]0=(16)(0)^{2}+c_{1} ext {, or } c_{1}=0[/latex] . Substituting this value into (2), we have

[latex]x=16 t^{2}[/latex] (3)

(c) We require t when x = 100. From (3) [latex]t=\sqrt{(100) /(16)}=2.5 sec[/latex].

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Eq. (7.5) : [latex]\frac{d v}{d t}=g[/latex]

Question 7.11: A body of mass 5 slugs is dropped from a height of 100 ft wi...

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