Question

If a ball is kicked with a perfectly horizontal velocity of 45.0 m/s and the ball was 1.5 m above the ground when the it was released, how far will the ball fly horizontally before it strikes the ground?

Group of answer choices

16.6 m

14 m

24.9 m

22.7 m

Answer 1

To find out how far the ball will fly horizontally before it strikes the ground, we first need to determine the time it takes for the ball to fall 1.5 meters vertically due to gravity.

We can use the following kinematic equation for vertical motion:

\[
d = \frac{1}{2} g t^2
\]

Where:
- \( d \) is the vertical distance (1.5 m),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \)),
- \( t \) is the time in seconds.

We can rearrange the equation to solve for \( t \):

\[
t^2 = \frac{2d}{g} \quad \Rightarrow \quad t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2 \times 1.5 \, m}{9.81 \, m/s^2}}
\]

Calculating \( t \):

\[
t = \sqrt{\frac{3}{9.81}} \approx \sqrt{0.3059} \approx 0.553 \, s
\]

Now that we have the time of flight, we can calculate the horizontal distance \( x \) traveled by the ball using the following equation:

\[
x = v_x \times t
\]

Where:
- \( v_x \) is the horizontal velocity (45.0 m/s),
- \( t \) is the time calculated earlier (approximately 0.553 s).

Now we substitute the values into the equation:

\[
x = 45.0 \, m/s \times 0.553 \, s \approx 24.885 \, m
\]

Rounding to one decimal place, the horizontal distance the ball will fly before it strikes the ground is approximately:

\[
\approx 24.9 \, m
\]

Therefore, the answer is **24.9 m**.