Question Statement

The height above ground for a ball dropped from an initial altitude of 122.5 m is given by $S(t)=122.5-4.9t^2$, where $S$ is measured in meters and $t$ in seconds.

(i) What is the instantaneous velocity at $t=\frac{1}{2}$ ?

(ii) At what time does the ball hit the ground?

(iii) What is the impact velocity?

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Background and Explanation

This problem applies the limit definition of the derivative to find instantaneous velocity from a position function. Understanding that velocity is the rate of change of position with respect to time allows us to analyze the ball's motion throughout its fall.

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Solution

Part (i): Instantaneous velocity at $t=\frac{1}{2}$

The instantaneous velocity at time $t_1$ is defined as the limit of the average velocity as the time interval approaches zero:

$V\left(t_1\right)={\lim }_{\Delta t\to 0}\frac{S\left(t_1+\Delta t\right)-S\left(t_1\right)}{\Delta t}$

Substituting the position function $S(t)=122.5-4.9t^2$:

$V\left(t_1\right)={\lim }_{\Delta t\to 0}\frac{\left(122.5-4.9{\left(t_1+\Delta t\right)}^2\right)-\left(122.5-4.9t_1^2\right)}{\Delta t}$

Expanding the squared term ${\left(t_1+\Delta t\right)}^2=t_1^2+2t_1\Delta t+\Delta t^2$:

$={\lim }_{\Delta t\to 0}\frac{122.5-4.9\left(t_1^2+\Delta t^2+2t_1\Delta t\right)-122.5+4.9t_1^2}{\Delta t}$

Distributing the $-4.9$ across the terms:

$={\lim }_{\Delta t\to 0}\frac{-4.9t_1^2-4.9\Delta t^2-9.8t_1\Delta t+4.8t_1^2}{\Delta t}$

Factoring out $\Delta t$ from the numerator:

$={\lim }_{\Delta t\to 0}\frac{\Delta t\left(-4.9\Delta t-9.8t_1\right)}{\Delta t}$

Canceling $\Delta t$ and evaluating the limit as $\Delta t\to 0$:

$=-4.9(0)-9.8t_1=-9.8t_1$

Thus, the velocity function is $V(t)=-9.8t$.

At $t_1=\frac{1}{2}$:

$V\left(\frac{1}{2}\right)=-9.8\times \frac{1}{2}=-4.9\mathrm{m}/\mathrm{s}$

Part (ii): Time when the ball hits the ground

When the ball strikes the ground, the height $S(t)=0$:

$\begin{array}{rl}S(t)& =0\\ 122.5-4.9t^2& =0\\ 122.5& =4.9t^2\\ t^2& =\frac{122.5}{4.9}\\ t^2& =25\\ t& =5\sec \end{array}$

(We take the positive root $t=5$ since time must be positive.)

Part (iii): Impact velocity

From the velocity function derived in part (i), $V(t)=-9.8t$.

At $t=5\sec$:

$\begin{array}{rl}V(5)& =-9.8\times 5\\ V(5)& =-49\mathrm{m}/\sec \end{array}$

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Key Formulas or Methods Used

• Limit definition of instantaneous velocity: $V(t)={\lim }_{\Delta t\to 0}\frac{S(t+\Delta t)-S(t)}{\Delta t}$
• Velocity function derived: $V(t)=-9.8t$ (where $-9.8$ represents acceleration due to gravity in $\mathrm{m}/{\mathrm{s}}^2$)
• Condition for ground impact: $S(t)=0$
• Kinematic relationship: Position function $S(t)=122.5-4.9t^2$ corresponds to initial height $122.5$ m and acceleration $-9.8\mathrm{m}/{\mathrm{s}}^2$

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Summary of Steps

1. Find velocity function: Apply the limit definition of the derivative to $S(t)$ to obtain $V(t)=-9.8t$
2. Calculate velocity at $t=\frac{1}{2}$: Substitute $t=0.5$ into $V(t)$ to get $-4.9\mathrm{m}/\mathrm{s}$ (negative indicates downward motion)
3. Find impact time: Set $S(t)=0$, solve $122.5-4.9t^2=0$ to get $t=5$ seconds
4. Calculate impact velocity: Substitute $t=5$ into $V(t)$ to get $-49\mathrm{m}/\mathrm{s}$