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Q101P

Expert-verifiedFound in: Page 91

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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?**

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$9.8\mathrm{m}/{\mathrm{s}}^{2}\left(\mathrm{downward}\right)$.

1) Initial velocity of the ball is ${\mathrm{V}}_{0}=7.00\mathrm{m}/\mathrm{s}$.

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

**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)

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,

$\mathrm{h}=\frac{{\mathrm{V}}_{0}^{2}-{\mathrm{V}}^{2}}{2\mathrm{g}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(7.00\mathrm{m}/\mathrm{s}\right)}^{2}-0}{2\times \left(9.8\mathrm{m}/{\mathrm{s}}^{2}\right)}\phantom{\rule{0ex}{0ex}}=2.50\mathrm{m}$

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

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,

${V}_{R}=7.00m/s-3.00m/s\phantom{\rule{0ex}{0ex}}=4.00m/s$

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

$h=\frac{{\left(4.00\mathrm{m}/\mathrm{s}\right)}^{2}-0}{2\times \left(9.81\mathrm{m}/{\mathrm{s}}^{2}\right)}\phantom{\rule{0ex}{0ex}}=0.82\mathrm{m}$

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

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$9.8\mathrm{m}/{\mathrm{s}}^{2}\left(\mathrm{downward}\right)$ .

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

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