Question Statement

A ball is thrown upward with a velocity of $40\text{ m/s}$.

1. Develop a differential equation representing the flow phenomenon.
2. Find the velocity of the ball after $1$ second.
3. Find the maximum height attained by the ball.

(Neglect air resistance and assume $g=9.8{\text{ m/s}}^2$)

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Background and Explanation

This problem uses Newton's Second Law to model motion under gravity. By expressing acceleration as the first derivative of velocity ($a=\frac{dv}{dt}$) and velocity as the first derivative of displacement ($v=\frac{dS}{dt}$), we can use integration to find the equations of motion.

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Solution

Part (a): Developing the Differential Equation

Since the ball is thrown upward and we are neglecting air resistance, the only acceleration acting on the ball is gravity ($g$), which acts in the opposite direction of the initial motion.

$\text{Acceleration }(a)=-g$

Since acceleration is the rate of change of velocity over time: $\frac{dv}{dt}=-g$

To find the velocity function, we separate the variables and integrate both sides: $dv=-gdt$ $\int dv=-g\int dt$ $v=-gt+A$

We use the initial condition to find the constant $A$. At $t=0$, the initial velocity $v=40\text{ m/s}$: $40=-g(0)+A$ $40=0+A⟹A=40$

Substituting $A$ back into equation (1), we get the velocity equation: $v=-gt+40$

Part (b): Velocity after 1 second

To find the velocity after 1 second, we substitute $t=1$ and $g=9.8{\text{ m/s}}^2$ into equation (2):

$v=-(9.8)(1)+40$ $v=-9.8+40$ $v=30.2\text{ m/sec}$

Part (c): Maximum Height Attained

To find the height, we first need the displacement equation. We know that velocity is the rate of change of displacement ($S$): $v=\frac{dS}{dt}=-gt+40$ $dS=(-gt+40)dt$

Integrating both sides: $\int dS=\int (-gt+40)dt$ $S=-g\frac{t^2}{2}+40t+B$

At $t=0$, the displacement $S=0$, so: $0=-g\frac{0^2}{2}+40(0)+B⟹B=0$

The displacement equation is: $S=-g\frac{t^2}{2}+40t$

At the maximum height, the velocity of the ball becomes zero ($v=0$). Using equation (2): $0=-gt+40$ $-gt=-40$ $t=\frac{40}{g}=\frac{40}{9.8}$ $t\approx 4.08\text{ sec}$

Now, substitute $t=4.08$ and $g=9.8$ into the displacement equation (4) to find the maximum height: $S=-9.8\frac{(4.08{)}^2}{2}+40(4.08)$ $S=-(4.9)(16.6464)+163.2$ $S=-81.56+163.2$ $S=81.64\text{ m}$

Hence, the maximum height is $81.6\text{ m}$.

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Key Formulas or Methods Used

• Definition of Acceleration: $a=\frac{dv}{dt}$
• Definition of Velocity: $v=\frac{dS}{dt}$
• Integration Rule: $\int t^{n}dt=\frac{t^{n+1}}{n+1}$
• Initial Value Problem: Using $t=0$ conditions to solve for constants of integration ($A$ and $B$).
• Physics Principle: At maximum height, instantaneous velocity $v=0$.

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Summary of Steps

1. Define the differential equation $\frac{dv}{dt}=-g$ based on gravity.
2. Integrate the acceleration to find the velocity function $v(t)$, using the initial velocity $40\text{ m/s}$ to solve for the constant.
3. Calculate $v(1)$ by plugging $t=1$ into the velocity function.
4. Integrate the velocity function to find the displacement function $S(t)$.
5. Determine the time $t$ when $v=0$ (the moment the ball reaches its peak).
6. Substitute that time into $S(t)$ to find the maximum height.