The angle of projection of a particle is measured from the vertical axis as $ \phi $ and the maximum height reached by the particle is $ h_m $. Here $ h_m $ as function of $ \phi $ can be presented as
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The Correct Option is C
Approach Solution - 1
\( H_{max} = \frac{u^2 \cos^2 \phi}{2g} \)
Approach Solution -2
To solve this problem, we need to determine the relationship between the maximum height \(h_m\) reached by a projectile and its angle of projection \(\phi\) with the vertical axis. Let's break down the problem:
The maximum height \(h_m\) of a projectile is given by the formula:
\(h_m = \frac{(v_0 \sin \theta)^2}{2g}\)
Here, \(v_0\) is the initial velocity, \(\theta\) is the angle of projection with the horizontal, and \(g\) is the acceleration due to gravity.
Since the angle of projection \(\phi\) is with the vertical, we relate it to the angle \(\theta\) by:
\(\theta = 90^\circ - \phi\)
Substituting \(\theta = 90^\circ - \phi\) into the maximum height formula, we get:
\(h_m = \frac{(v_0 \sin(90^\circ - \phi))^2}{2g}\)
We know that \(\sin(90^\circ - \phi) = \cos \phi\). Thus, the expression becomes:
\(h_m = \frac{(v_0 \cos \phi)^2}{2g}\)
Simplifying this, we have:
\(h_m = \frac{v_0^2 \cos^2 \phi}{2g}\)
This matches the given correct answer. Therefore, the relationship between the maximum height \(h_m\) and the angle \(\phi\) is:
Thus, the correct answer is that \(h_m\) can be expressed as:
\(h_m = \frac{v_0^2 \cos^2 \phi}{2g}\)
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