Question

In Fig. 4-55, a ball is shot directly upward from the ground with an initial speed of ${\mathbf{V}}_{\mathbf{0}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}\mathbf{s}$. Simultaneously, a construction elevatorcab begins to move upward from the ground with a constant speed of ${\mathbf{V}}_c\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}\mathbf{s}$. What maximum height does the ball reach relative to (a) the ground and (b) the cab floor? At what rate does the speed of the ball change relative to (c) the ground and (d) the cab floor?

Short answer

A) Maximum height of the ball relative to the ground is $2.50m$.

B) Maximum height of the ball relative to the cab floor$0.82\mathrm{m}$.

C) Rate of speed change of ball relative to ground$9.8\mathrm{m}/{\mathrm{s}}^2\left(\mathrm{downward}\right)$.

D) Rate of speed change of ball relative to cab floor$9.8\mathrm{m}/{\mathrm{s}}^2\left(\mathrm{downward}\right)$.

Step-by-step solution

Given data

1) Initial velocity of the ball is ${\mathrm{V}}_0=7.00\mathrm{m}/\mathrm{s}$.

2) Speed of construction elevator is ${\mathrm{V}}_c=3.00\mathrm{m}/\mathrm{s}$.

To understand the concept of relative velocity and gravitational force

The cab floor and the ball both are under the influence of gravitational force. So, we can use the equations of constant acceleration for their motion. Using the kinematic equations in terms of initial, final velocity, acceleration, and displacement, we can find the height the ball can reach.

Formula:

$V^2=V_0^2+2a\left(X-X_0\right)$ …(i)

(a) Calculate the maximum height the ball reaches relative to the ground

When ball reach to the maximum height (h), its final velocity (v) will be zero. The acceleration on the ball is the gravitational acceleration (g) and it is acting in a downward direction. If we assume the upward direction as positive, the gravitational acceleration is considered negative. Therefore, the equation (i) becomes,

$V^2={V^2}_0-2gh$

Rearranging this for height h we get,

$$\begin{gathered} \mathrm{h}=\frac{{\mathrm{V}}_0^2-{\mathrm{V}}^2}{2\mathrm{g}} \\ =\frac{{\left(7.00\mathrm{m}/\mathrm{s}\right)}^2-0}{2\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)} \\ =2.50\mathrm{m} \end{gathered}$$

Therefore, the maximum height of the ball relative to the ground is$2.50m$

(b) Calculate the maximum height the ball reaches relative to the cab floor

The cab floor is moving upward with a velocity${\mathrm{V}}_c=3.00\mathrm{m}/\mathrm{s}$

So, the relative velocity of a ball is,

$$\begin{gathered} V_R=7.00m/s-3.00m/s \\ =4.00m/s \end{gathered}$$

Following the similar procedure as part a), we can find the maximum height as

$$\begin{gathered} h=\frac{{\left(4.00\mathrm{m}/\mathrm{s}\right)}^2-0}{2\times \left(9.81\mathrm{m}/{\mathrm{s}}^2\right)} \\ =0.82\mathrm{m} \end{gathered}$$

The maximum height the ball reaches relative to the cab floor is$0.82\mathrm{m}$

(c) Calculate the rate at which the speed of the ball changes relative to the ground

Ball has only gravitational force acting on it, so the acceleration of a ball or its rate of speed change is equal to the acceleration due to gravity that is$9.8\mathrm{m}/{\mathrm{s}}^2\left(\mathrm{downward}\right)$ .

(d) Calculate the rate at which the speed of the ball changes relative to the cab floor

The cab floor is moving with constant velocity, so it will also have similar acceleration as that of ball that is $9.8\mathrm{m}/{\mathrm{s}}^2\left(\mathrm{downward}\right)$.