MAXIMUM HEIGHT DERIVED GOAL Find the maximum height of a thrown projectile using symbols. PROBLEM Refer to Example 2.8. Use symbolic manipulation to find (a) the time tmax it takes the ball to reach its maximum height and (b) an expression for the maximum height that doesn't depend on time. Answers

Question

MAXIMUM HEIGHT DERIVED

GOAL Find the maximum height of a thrown projectile using symbols.

PROBLEM Refer to Example 2.8. Use symbolic manipulation to find (a) the time [latex]t_{\max }[/latex] it takes the ball to reach its maximum height and (b) an expression for the maximum height that doesn't depend on time. Answers should be expressed in terms of the quantities [latex]v_{0}, g[/latex], and [latex]y_{0}[/latex] only.

STRATEGY When the ball reaches its maximum height, its velocity is zero, so for part (a) solve the kinematics velocity equation for time [latex]t[/latex] and set [latex]v=0[/latex]. For part (b), substitute the expression for time found in part (a) into the displacement equation, solving it for the maximum height.

Accepted Answer

(a) Find the time it takes the ball to reach its maximum height.

Write the velocity kinematics equation:

[latex]v=a t+v_{0} [/latex]

Move [latex]v_{0}[/latex] to the left side of the equation:

[latex]v-v_{0}=a t [/latex]

Divide both sides by [latex]a[/latex] :

[latex]\frac{v - v_{0}}{a}=\frac{\cancel{a} t}{\cancel{a}}=t [/latex]

Turn the equation around so that [latex]t[/latex] is on the left and substitute [latex]v=0[/latex], corresponding to the velocity at maximum height:

[latex] ext { (1) } t=\frac{-v_{0}}{a}[/latex]

Replace [latex]t[/latex] by [latex]t_{\max }[/latex] and substitute [latex]a=-g[/latex] :

[latex] ext { (2) } t_{\max }=\frac{v_{0}}{g}[/latex]

(b) Find the maximum height.

Write the equation for the position [latex]y[/latex] at any time:

[latex]y= y_{0}+v_{0} t+\frac{1}{2} a t^{2} [/latex]

Substitute [latex]t=v_{0} / a[/latex], which corresponds to the time it takes to reach [latex]y_{\max }[/latex], the maximum height:

[latex]y_{\max } =y_{0}+v_{0}\left(\frac{-v_{0}}{a} ight)+\frac{1}{2} a\left(\frac{-v_{0}}{a} ight)^{2} [/latex]

[latex]=y_{0} -\frac{v_{0}^{2}}{a}+\frac{1}{2} \frac{v_{0}^{2}}{a}[/latex]

Combine the last two terms and substitute [latex]a= - g[/latex] :

(3) [latex]y_{\max }=y_{0}+\frac{v_{0}^{2}}{2 g}[/latex]

REMARKS Notice that [latex]g=+9.8 \mathrm{~m} / \mathrm{s}^{2}[/latex], so the second term is positive overall. Equations (1) [latex]-(3)[/latex] are much more useful than a numerical answer because the effect of changing one value can be seen immediately. For example, doubling the initial velocity [latex]v_{0}[/latex] quadruples the displacement above the point of release. Notice also that [latex]y_{ ext {max }}[/latex] could be obtained more readily from the timeindependent equation, [latex]v^{2}-v_{0}^{2}=2 a \Delta y[/latex].

Question 2.9: MAXIMUM HEIGHT DERIVED GOAL Find the maximum height of a thr...

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