Question

You throw a baseball at an initial speed of 15.0 m/s at an angle of 30° with respect to the horizontal. What would the ball’s initial speed have to be at 30° on a planet that has twice the acceleration of gravity as Earth to achieve the same range? Consider launch and impact on a horizontal surface.

a) 10.6 m/s
b) 12.1 m/s
c) 15.0 m/s
d) 18.3 m/s

Answer 1

The ball's initial speed at 30° on a planet with twice the acceleration of gravity as Earth to achieve the same range would be
`\(18.3 \, \text{m/s}\)` (Option D).

Step-by-step explanation:

To determine the required initial speed, we can use the range formula for projectile motion:

`\[ R = (v_0^2 \sin(2\theta))/(g) \]`

where:

- ( R ) is the range,

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

-
`\( \theta \)` is the launch angle,

- ( g ) is the acceleration due to gravity.

Since we want the range to remain the same, we set the two scenarios equal to each other:

`\[ (v_(0_1)^2 \sin(2\theta_1))/(g_1) = (v_(0_2)^2 \sin(2\theta_2))/(g_2) \]`

Given that
`\(g_2 = 2g_1\)` for the other planet, and
`\( \theta_1 = \theta_2 = 30^\circ \)` (same launch angle), we can simplify the equation:

`\[ (v_(0_1)^2)/(g_1) = (v_(0_2)^2)/(2g_1) \]`

Now, solving for
`\(v_(0_2)\)`:

\
`[ v_(0_2) = √(2) \cdot v_(0_1) \]`

Substituting the given initial speed
`(\(v_(0_1) = 15.0 \, \text{m/s}\))`, we get:

`\[ v_(0_2) = √(2) \cdot 15.0 \, \text{m/s} \approx 21.2 \, \text{m/s} \]`

However, this is the speed required for the same range. To find the speed at the same launch angle, we need to find the horizontal component, which is
`\(v_(0_x) = v_(0) \cos(\theta)\)`:

\
`[ v_(0_2x) = √(2) \cdot v_(0_1) \cos(30^\circ) \]`

`\[ v_(0_2x) = √(2) \cdot 15.0 \, \text{m/s} \cdot (√(3))/(2) \approx 18.3 \, \text{m/s} \]`

Therefore, the correct answer is
`\(18.3 \, \text{m/s}\)` (Option D).