Question

What is the equation to calculate the maximum height reached by an object projected upwards at an angle of 60 degrees with an initial velocity of 25 meters per second?

Answer 1

To calculate the maximum height reached by an object projected upwards at an angle, you can use the following equation derived from the kinematic equations of motion. The maximum height \( H \) can be calculated using:

\[
H = \frac{(v_{0y})^2}{2g}
\]

where:
- \( v_{0y} \) is the initial vertical component of the velocity,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

First, we need to find the initial vertical component of the velocity \( v_{0y} \) for the given initial velocity \( v_0 = 25 \, \text{m/s} \) and an angle \( \theta = 60^\circ \).

The vertical component can be calculated as:

\[
v_{0y} = v_0 \cdot \sin(\theta)
\]

Calculating \( v_{0y} \):

\[
v_{0y} = 25 \cdot \sin(60^\circ)
\]

Using \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):

\[
v_{0y} = 25 \cdot \frac{\sqrt{3}}{2} \approx 25 \cdot 0.866 \approx 21.65 \, \text{m/s}
\]

Now, substituting \( v_{0y} \) into the height equation:

\[
H = \frac{(21.65)^2}{2 \cdot 9.81}
\]

Calculating \( (21.65)^2 \):

\[
(21.65)^2 \approx 468.6225
\]

Now, calculate \( 2g \):

\[
2g \approx 2 \cdot 9.81 \approx 19.62
\]

Therefore, computing \( H \):

\[
H \approx \frac{468.6225}{19.62} \approx 23.87 \, \text{meters}
\]

So, the maximum height reached by the object is approximately **23.87 meters**.