Question

A boy can throw a ball 10 m high in his full capacity. how far away (in horizontal direction) can he throw the same ball _____ (in metre)?

a. 20
b. 20√2
c. 10
d. 10√2

Answer 1

The boy can throw the ball approximately 20 meters away when throwing at a 45-degree angle, (A)

How to find distance?

The formula for the maximum height h reached by a projectile is:

`\[ h = (v^2 \sin^2(\theta))/(2g) \]`

For a vertical throw,
`\( \theta = 90^\circ \)`, so
`\( \sin^2(\theta) = 1 \)`.

Rearranging for v:

`\[ v = √(2gh) \]`

Substituting
`\( g = 9.81 \, \text{m/s}^2 \)` and
`\( h = 10 \, \text{m} \)`, find:

`\[ v = √(2 * 9.81 * 10) \approx 14.01 \, \text{m/s} \]`

The range formula for a projectile is:

`\[ R = (v^2 \sin(2\theta))/(g) \]`

For
`\( \theta = 45^\circ \)`,
`\( \sin(2 * 45^\circ) = \sin(90^\circ) = 1 \)`.

Therefore, the range formula simplifies to:

`\[ R = (v^2)/(g) \]`

Using the initial velocity
`\( v = 14.01 \, \text{m/s} \)`:

`\[ R = ((14.01)^2)/(9.81) \approx 20 \, \text{m} \]`

So, with the initial velocity of about 14.01 m/s (needed to reach 10 meters high), the boy can throw the ball approximately 20 meters away when throwing at a 45-degree angle.