Q82P - University Physics with Modern Physics

Question

A ball is thrown straight up from the ground with speed ${\mathbf{v}}_{\mathbf{0}}$. At the same instant, a second ball is dropped from rest from a height H, directly above the point where the first ball was thrown upward. There is no air resistance. (a) Find the time at which the two balls collide. (b) Find the value of H in terms of ${\mathbf{v}}_{\mathbf{0}}$ and g such that at the instant when the balls collide, the first ball is at the highest point of its motion.

Step-by-Step Solution

Answer

Answer is Missing in the Document.

Step 1: Identification of the given data

The given data can be listed below as,

 

• The initial velocity of the ball when it was thrown from the ground is, $v_0$.
• Height from which the second ball is thrown downward is, H

Step 2: Explanation of the second equation of motion

In the effect of uniform acceleration, the displacement of a particular item can be obtained by the second equation of motion. It is expressed as follows,

 

$\mathit{s}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$                                                                                                               

 

Here, s is the distance travelled by a specific item, u is the initial velocity of the item, a is the acceleration of the item and t is the time taken to travel by the item.

Step 3: Determination of the time of collision of two balls

(a)

 

Write the expression for the height attained by first ball.

 

$s_1=V_{0}t-\frac{1}{2}g{t}^2$                                                                                                         …(1)

 

Here, g is the acceleration due to gravity.

 

Write the expression for the height attained by second ball.

 

$S_2=H-\frac{1}{2}g{t}^2$

 

As time of collision is to be obtained, so equate both the heights.

 

$$\begin{gathered} s_1=s_2 \\ {v}_{0}t-\frac{1}{2}g{t}^2=H-\frac{1}{2}g{t}^2 \\ {v}_{0}t=H \\ t=\frac{H}{v_0} \end{gathered}$$

 

Thus, the time of collision of two balls is $\frac{H}{v_0}$.

Step 4: Determination of the value of H when the first ball is at the highest position when both balls collide

 

(b)

 

Write the expression for the velocity of first ball when it reaches its highest position.

 

$v_1=v_0-gt$

 

Here, $v_1$ is the final velocity of the first ball and that is zero.

 

Substitute the values and solve.

 

$$\begin{gathered} 0=v_0-gt \\ gt=v_0 \\ t=\frac{v_0}{g} \end{gathered}$$

 

Substitute the value of time from the above expression in equation (1).

 

$$\begin{gathered} s_1=v_0\left(\frac{v_0}{g}\right)-\frac{1}{2}g{\left(\frac{v_0}{g}\right)}^2 \\ =\frac{{\left(v_0\right)}^2}{g}-\frac{{\left(v_0\right)}^2}{2g} \\ =\frac{{\left(v_0\right)}^2}{2g} \end{gathered}$$

 

Write the expression for the distance covered by second ball at the same time that is $t=\frac{V_0}{g}$.

 

$$\begin{gathered} H-s_2=H-\left(H-\frac{1}{2}g{t}^2\right) \\ =\frac{1}{2}g{t}^2 \end{gathered}$$

 

Substitute the value of time in the above expression.

 

$$\begin{gathered} H-s_2=\frac{1}{2}g{\left(\frac{v_0}{g}\right)}^2 \\ =\frac{{\left(v_0\right)}^2}{2g} \end{gathered}$$

 

Write the expression for the value of H, at time $t=\frac{V_0}{g}$.

 

$$\begin{gathered} H=\frac{{\left(v_0\right)}^2}{2g}+\frac{{\left(v_0\right)}^2}{2g} \\ =\frac{{\left(v_0\right)}^2}{g} \end{gathered}$$

 

Thus, the value of H when the first ball is at the highest position when both balls collide is $\frac{{\left(v_0\right)}^2}{g}$