Question

a ball is rolled of the edge of a table with a
horizontal relocity of 10 m/s .What will be the ball's
velocity 0.10s later?

Answer 1

Answer:

Approximately $10\; \rm m \cdot s^{-1}$ at $5.6^\circ$ below the horizon.

• Horizontal component of velocity: $10\; \rm m \cdot s^{-1}$.
• Vertical component of velocity: $0.981\; \rm m\cdot s^{-1}$ (downwards.)

(Assumption: air resistance on the ball is negligible; $g = 9.81\; \rm m \cdot s^{-2}$.)

Explanation:

Assume that the air resistance on the ball is negligible. The horizontal component of the velocity of the ball would stay the same at $10\; \rm m \cdot s^{-1}$ until the ball reaches the ground.

On the other hand, the vertical component of the ball would increase (downwards) at a rate of $g = 9.81\; \rm m\cdot s^{-2}$ (where $g$ is the acceleration due to gravity.) In $0.1\; \rm s$, the vertical component of the velocity of this ball would have increased by $9.81\; \rm m \cdot s^{-2} \times 0.10\; \rm s = 0.981\; \rm m \cdot s^{-1}$.

However, right after the ball rolled off the edge of the table, the vertical component of the velocity of this ball was $0\; \rm m\cdot s^{-1}$. Hence, $0.10\; \rm s$ after the ball rolled off the table, the vertical component of the velocity of this ball would be $0\; \rm m \cdot s^{-1} + 0.981\; \rm m\cdot s^{-1} = 0.981\; \rm m \cdot s^{-1}$.

Calculate the magnitude of the velocity of this ball. Let $v_{x}$ and $v_{y}$ and denote the horizontal and vertical component of the velocity of this ball, respectively. The magnitude of the velocity of this ball would be $\displaystyle \sqrt{{v_x}^{2} + {v_y}^{2}}$.

At $0.10\; \rm s$ after the ball rolled off the table, $v_x = 10\; \rm m \cdot s^{-1}$ while $v_y = 0.981\; \rm m \cdot s^{-1}$. Calculate the magnitude of the velocity of the ball at this moment:

$\begin{aligned} \| v \| &= \sqrt{{v_x}^{2} + {v_y}^{2}} \\ &= \sqrt{\left(10\; \rm m \cdot s^{-1}\right)^{2} + \left(0.981\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 10.0\; \rm m\cdot s^{-1}\end{aligned}$.

Calculate the angle between the horizon and the velocity of the ball (a vector) at that moment. Let $\theta$ denote that angle.

$\displaystyle \tan \theta = \frac{\text{rise}}{\text{run}}$.

For the vector representing the velocity of this ball:

$\displaystyle \tan \theta = \frac{0.981\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-1}} = 0.0981$.

Calculate the size of this angle:

$\theta = \arctan 0.0981 \approx 5.62^\circ$.

Notice that the vertical component of the velocity of this ball at that moment points downwards (towards the ground.) Hence, the corresponding velocity should point below the horizon.