Q89CP - University Physics with Modern Physics

Question

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{s}$ later. Ignore air resistance. (a) If the height of the building is $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{m}$, what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed ${\mathbf{v}}_{\mathbf{0}}$ of the first ball be given and treat the height of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) ${\mathbf{v}}_{\mathbf{0}\mathbf{ }}\mathbf{is}\mathbf{ }\mathbf{9}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{ }$? (c) If ${\mathbf{v}}_{\mathbf{0}}$is greater than some value${\mathbf{v}}_{\mathbf{max}}$, no value of exists that allows both balls to hit the ground at the same time. Solve for ${\mathbf{v}}_{\mathbf{max}}$. The value ${\mathbf{v}}_{\mathbf{max}}$ has a simple physical interpretation. What is it? (d) If ${\mathbf{v}}_{\mathbf{0}}$ is less than some value ${\mathbf{v}}_{\mathbf{min}}$, no value of exists that allows both balls to hit the ground at the same time. Solve for ${\mathbf{v}}_{\mathbf{min}}$. The value ${\mathbf{v}}_{\mathbf{min}}$also has a simple physical interpretation. What is it?

Step-by-Step Solution

Answer

a)The initial speed of the first ball when both the balls hit the ground is.

 

The graph has been drawn below-

b)The height of the building at different velocities are $0.411 \mathrm{m} \mathrm{and} 1152.04 \mathrm{m}$

 respectively.

c)The value of ${\mathrm{v}}_{\max } \mathrm{is} 9.8 \mathrm{m}/\mathrm{s}$.

d)The value of ${\mathrm{v}}_{\min } \mathrm{is} 4.9 \mathrm{m}/\mathrm{s}$.

Step 1: Identification of given data

• The height of the building is, $\mathrm{h}=20.00 \mathrm{m}$.
• The initial speed of the ball, ${\mathrm{v}}_0$.

• The time at which second ball is dropped is,${\mathrm{t}}_0=1.00\mathrm{s}$. 

Step 2: Concept of maximum speed

When an athlete reaches their top speed, they are said to have achieved their maximum potential.

Athletes' acceleration varies as they vary the magnitude of their movement, since velocity has magnitude and direction both.

Step 3: Determine the initial speed of the first ball

a)

As, $+Y$is upward, then acceleration of each ball is,

$a_y=-g$

The height equation of two ball will be,

$\begin{array}{l}{\mathrm{y}}_1=\mathrm{h}+{\mathrm{v}}_0^{}{\mathrm{t}}_1-\frac{1}{2}\mathrm{g}{{\mathrm{t}}_1}^2\\ {\mathrm{y}}_2=\mathrm{h}-\frac{1}{2}{\mathrm{g}}_2^2\end{array}$                             …1)

Here, ${\mathrm{y}}_1 \mathrm{and} {\mathrm{y}}_2$are the displacement of the first and the second ball, h is the height, ${\mathrm{v}}_0$is the initial speed, $t_1$is the initial time and $t_2$is the final time and g is the acceleration due to gravity.

As the second ball has been dropped after $1\mathrm{s}$, then the final time becomes,

$t_2=t_1-1$

Substituting the values in the equation 1),

$y_2=h-\frac{1}{2}g{\left(t_1-1\right)}^2$

Here, $t_0=1.0s$

When two ball hits the ground$y_1=y_2=0$

$h+V_0{t}_1-\frac{1}{2}{g}_1^2=h-\frac{1}{2}9{\left(t_1-1\right)}^2$

Solving the above equation,

$$\begin{gathered} v_0{t}_1-\frac{1}{2}g{t}_1^2=-\frac{1}{2}g{t}_1^2+g{t}_1^{}-\frac{1}{2}g \\ v_0{t}_1=g{t}_1^{}--\frac{1}{2}g \\ t_1=\frac{g}{2\left(g-v_0\right)} \end{gathered}$$

Substituting the above value in the equation 1), 

$$\begin{gathered} 0=h+v_0\left[\frac{g}{2\left(g-v_0\right)}\right]-\frac{1}{2}g{\left[\frac{g}{2\left(g-v_0\right)}\right]}^2 \\ h=\frac{1}{2}g\left[\frac{g^2-4v_0\left(g-v_0\right)}{4{\left(g-v_0\right)}^2}\right] \\ =\frac{1}{2}g\left[\frac{{\left(\frac{1}{2}g\right)}^2-v_0{g}_0+v_0^2}{{\left(g-v_0\right)}^2}\right] \end{gathered}$$

Hence, further as,

$\begin{array}{l}h=\frac{1}{2}g,\frac{{\left({\displaystyle \frac{1}{2}}g-v_0\right)}^2}{\left(g-v_0\right)}\\ v_0^2\left(2h-g\right)-v_0\left(4gh-g^2\right)+g^2\left(2h-\frac{g}{4}\right)=0\end{array}$

Substituting the values of h and g in the above equation,

$30.2v_0^2-678.96v_0+3606.302=0$

Solving the equation, the value of $v_0$becomes $8.17 \mathrm{m}/\mathrm{s} \mathrm{and} 14.6 \mathrm{m}/\mathrm{s}$

Here, the value of ${\mathrm{v}}_0 \mathrm{is} 8.17 \mathrm{m}/\mathrm{s}$. The reason that the second value is not taken is that after ,the first ball is rising and the second ball started to fall.

Thus, the initial speed of the first ball when both the balls hit the ground is$8.17 \mathrm{m}/\mathrm{s}$.

The graph has been drawn below-

                      

Step 4:Determine the height of the building

The equation of the height can be expressed as,

$h=\frac{1}{2}g\left[\frac{{\left(\frac{1}{2}g-v_0\right)}^2}{{\left(g-v_0\right)}^2}\right]$                                …2)

The height of the building in both cases can be evaluated,

Substituting the values in the equation 2),

       (I)  The height of the building when velocity of the ball is,$V_0=6.0\mathrm{m}/\mathrm{s}$

$$\begin{gathered} {\mathrm{h}}_1=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(\frac{4.9\mathrm{m}/\mathrm{s}-6.0\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/\mathrm{s}-6.0\mathrm{m}/\mathrm{s}}\right)}^2 \\ =0.411\mathrm{m} \end{gathered}$$

        (ii)  The height of the building when the velocity of ball is,$v_{0}9.5\mathrm{m}/\mathrm{s}$

$$\begin{gathered} {\mathrm{h}}_2=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(\frac{4.9\mathrm{m}/\mathrm{s}-9.5\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/\mathrm{s}-9.5\mathrm{m}/\mathrm{s}}\right)}^2 \\ =1152.04\mathrm{m} \end{gathered}$$

Thus, the height of the building at different velocities is$0.411\mathrm{m} \mathrm{and} 1152.04\mathrm{m}$ respectively.

Step 5: Evaluate v max   when   v 0 is greater than some value

c)

 

The height h approaches infinite when $v_0$approaches$9.8\mathrm{m}/\mathrm{s}$, which corresponds to a relative velocity approaching zero when the second ball is in the air. No matter what $v_0$ is, the first ball cannot catch the second one. Hence, the value of ${\mathrm{v}}_{\max }$is $9.8\mathrm{m}/\mathrm{s}$.

Thus, the value of ${\mathrm{v}}_{\max } \mathrm{is} 9.8\mathrm{m}/\mathrm{s}$.

Step 6: Evaluate the v min   when   v 0 is less than v min

d)

When the speed of ${\mathrm{v}}_0 \mathrm{reches} 4.9\mathrm{m}/\mathrm{s}$, the height decreases to zero. When the second ball leaves, the first ball will be nearer the top of the roof than it was when it was released. As soon as the first ball is launched, it will have already gone through the ceiling on its way down, meaning that the second ball will never be able to catch up. Hence, the value of ${\mathrm{v}}_{\min } \mathrm{is} 4.9\mathrm{m}/\mathrm{s}$.

Thus, the value of is${\mathrm{v}}_{\min } \mathrm{is} 4.9\mathrm{m}/\mathrm{s}$.