Question

What is the maximum height reached by a projectile launched at an angle of 60 degrees with an initial velocity of 20 meters per second?

Answer 1

To determine the maximum height reached by a projectile, we can use the following formula:

\[
H = \frac{v_y^2}{2g}
\]

where:
- \( H \) is the maximum height,
- \( v_y \) is the initial vertical velocity,
- \( g \) is the acceleration due to gravity (approx. \( 9.81 \, \text{m/s}^2 \)).

First, we need to find the initial vertical velocity \( v_y \). The initial velocity \( v_0 \) is given as 20 m/s, and we can find \( v_y \) using the angle of projection \( \theta = 60^\circ \):

\[
v_y = v_0 \sin(\theta)
\]

Calculating \( v_y \):

\[
v_y = 20 \sin(60^\circ)
\]

We know that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \). Therefore:

\[
v_y = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s}
\]

Now we can substitute \( v_y \) into the height formula:

\[
H = \frac{(10\sqrt{3})^2}{2 \cdot 9.81}
\]

Calculating \( (10\sqrt{3})^2 \):

\[
(10\sqrt{3})^2 = 100 \cdot 3 = 300
\]

Now substituting this into the height formula:

\[
H = \frac{300}{2 \cdot 9.81} = \frac{300}{19.62} \approx 15.29 \text{ meters}
\]

Thus, the maximum height reached by the projectile is approximately \( \boxed{15.29} \) meters.