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

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)`.)

Step-by-step 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) * 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 = (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.