Q. 6 In our derivation of the positio... [FREE SOLUTION]

Chapter 11: Q. 6 (page 898)

In our derivation of the position function $r (t) = <(||v_{0}|| \cos 胃) t, - \frac{1}{2} g t^{2} + (||v_{0}|| \sin 胃) t + h>$ for the motion of a projectile, we ignored air resistance and wind effects. In Exercises 4 and 5 we ask you to think about how these effects would change the model.
A projectile reaches its maximum height H at time $t_{0} > 0$ . What fraction of H does it reach at time $\frac{1}{2} t_{0}$ ?

Short Answer

Expert verified

The fraction of H it reach at time $\frac{1}{2} t_{0}$ is $(\frac{4 (||v_{0}|| \sin 胃) - g t_{0}}{8 (||v_{0}|| \sin 胃) - 4 g t_{0}}) H .$

Step by step solution

Step 1. Given Information.

Given is a model of projectile motion with coordinates:

$r (t) = <(||v_{0}|| \cos 胃) t, - \frac{1}{2} g t^{2} + (||v_{0}|| \sin 胃) t + h>$

and $t = 0.5 t_{0} .$

Step 2. Calculation of the vertical component.

$A t t = 0 i t i s a t g r o u n d l e v e l s o h_{1} = 0, N o w a t t i m e t = 0.5 t_{0} i t w i l l b e a t h_{2} = - \frac{1}{8} g {t_{0}}^{2} + (||v_{0}|| \sin 胃) \frac{t_{0}}{2} + h_{1} h_{2} = - \frac{1}{8} g {t_{0}}^{2} + (||v_{0}|| \sin 胃) \frac{t_{0}}{2} . A n d w e k n o w t h a t H = - \frac{1}{2} g {t_{0}}^{2} + (||v_{0}|| \sin 胃) t_{0} . S o, \frac{h_{2}}{H} = \frac{- \frac{1}{8} g {t_{0}}^{2} + (||v_{0}|| \sin 胃) \frac{t_{0}}{2}}{- \frac{1}{2} g {t_{0}}^{2} + (||v_{0}|| \sin 胃) t_{0}} = \frac{1}{4} \frac{- \frac{1}{2} g {t_{0}}^{2} + 2 (||v_{0}|| \sin 胃) t_{0}}{- \frac{1}{2} g {t_{0}}^{2} + (||v_{0}|| \sin 胃) t_{0}} h_{2} = (\frac{4 (||v_{0}|| \sin 胃) - g t_{0}}{8 (||v_{0}|| \sin 胃) - 4 g t_{0}}) H .$

$S o a t t = 0.5 t_{0} i t w i l l b e a t h_{2} = (\frac{4 (||v_{0}|| \sin 胃) - g t_{0}}{8 (||v_{0}|| \sin 胃) - 4 g t_{0}}) H .$

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Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Calculation of the vertical component.

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