Question

A ball rolls off a horizontal table top with a speed of 1.7m/s and strike the flow in0.45s compute the ff.
A. The height of the table above the floor
B. The speed of the ball when it strikes the ball

Answer 1

Answer on Question 79357, Physics, Other

Question:

A ball rolls off a horizontal table top with a speed of $1.7\ \mathrm{m/s}$ and strike the floor in $0.45\ \mathrm{s}$. Compute the following:

a) The height of the table above the floor.

b) The speed of the ball when it strikes the floor.

Solution:

a) We can find the height of the table above the floor from the kinematic equation:

$$y = v_{0y}t + \frac{1}{2}gt^2,$$

here, $y$ is the height of the table above the floor, $v_{0y} = 0$ is the vertical component of the initial speed of the ball, $g = 9.8\ \mathrm{m/s^2}$ is the acceleration due to gravity and $t$ is the time.

Then, we get:

$$y = \frac{1}{2}gt^2 = \frac{1}{2} \cdot 9.8\ \frac{\mathrm{m}}{\mathrm{s^2}} \cdot (0.45\ \mathrm{s})^2 = 0.99\ \mathrm{m}.$$

b) Let's first find the vertical component of the final speed of the ball:

$$v_{fy} = v_{0y} + gt = 9.8\ \frac{\mathrm{m}}{\mathrm{s^2}} \cdot 0.45\ \mathrm{s} = 4.41\ \frac{\mathrm{m}}{\mathrm{s}}.$$

Finally, we can find the final speed of the ball when it strikes the floor from the Pythagorean theorem: $v_{f} = \sqrt{v_{fx}^{2} + v_{fy}^{2}}$, here, $v_{fx} = 1.7\ \mathrm{m/s}$ is the horizontal component of the final speed of the ball, $v_{fy} = 4.41\ \mathrm{m/s}$ is the vertical component of the final speed of the ball.

Then, we get:

$$v_{f} = \sqrt{v_{fx}^{2} + v_{fy}^{2}} = \sqrt{\left(1.7\ \frac{\mathrm{m}}{\mathrm{s}}\right)^{2} + \left(4.41\ \frac{\mathrm{m}}{\mathrm{s}}\right)^{2}} = 4.72\ \frac{\mathrm{m}}{\mathrm{s}}.$$

Solution:

a) $y = 0.99\ \mathrm{m}$.

b) $v_{f} = 4.72\ \frac{\mathrm{m}}{\mathrm{s}}$.

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