Question

The boy on the tower throws a ball 20 meters downrange as shown.

What is his pitching speed? (Take g = 9.8 m/s²)
Show work clearly and completely (including formulas, calculations, units, etc. as done in class.)

Answer 1

The problem states that the ball is thrown 20 meters downrange. Since the ball is thrown horizontally, we can assume that the initial vertical velocity is zero. The only force acting on the ball is gravity, which causes it to accelerate downwards at a rate of 9.8 m/s² 1.

To calculate the pitching speed, we need to find the time it takes for the ball to travel 20 meters horizontally. Unfortunately, this information is not provided in the problem statement. Therefore, we cannot calculate the pitching speed.

Answer 2

The pitching speed of the ball is
`\(20 \, \text{m/s}\).`

To find the pitching speed of the ball, we can use the following kinematic equation for projectile motion:

`\[ h = (1)/(2) g t^2 \]`

where:

- h is the initial vertical displacement (4.9 m, the height of the tower),

- g is the acceleration due to gravity (9.8 m/s²),

- t is the time of flight.

We can rearrange this equation to solve for t:

`\[ t = \sqrt{(2h)/(g)} \]`

Substitute the given values:

`\[ t = \sqrt{\frac{2 * 4.9 \, \text{m}}{9.8 \, \text{m/s}^2}} \]`

`\[ t = √(1) \, \text{s} = 1 \, \text{s} \]`

Now, we can use the formula for horizontal displacement in projectile motion:

`\[ \text{Range} = v_0 \cos(\theta) * t \]`

where:

- Range is the horizontal distance (20 m),

-
`\(v_0\)` is the initial velocity (pitching speed),

-
`\(\theta\)` is the angle of projection (assumed to be horizontal),

- t is the time of flight.

Rearrange the formula to solve for
`\(v_0\)`:

`\[ v_0 = \frac{\text{Range}}{t} \]`

Substitute the known values:

`\[ v_0 = \frac{20 \, \text{m}}{1 \, \text{s}} = 20 \, \text{m/s} \]`

Therefore, the pitching speed of the ball is
`\(20 \, \text{m/s}\).`