Question

a stone is thrown upwards with a velocity of 40m/s. if the maximum height attained is 20m. what is the angle of projection

Answer 1

Answer on Question #58248-Physics-Mechanics-Relativity

A stone is thrown upwards with a velocity of $40\mathrm{m/s}$ if the maximum height attained is $20\mathrm{m}$. What is the angle of projection?

Solution

The maximum height of projectile can be obtained by the formula:

$$h_{\max} = \frac{v_0^2}{2g} \sin^2 \theta \rightarrow \sin^2 \theta = \frac{2gh_{\max}}{v_0^2},$$

where $\theta$ is the angle of projection, $v_0$ is initial velocity of projectile, $g$ is the acceleration due to the gravity.

Thus, the angle of projection is

$$\theta = \sin^{-1} \sqrt{\frac{2gh_{\max}}{v_0^2}} = \sin^{-1} \sqrt{\frac{2 \cdot 10 \cdot 20}{40^2}} = 30{}^\circ$$

Answer: $30{}^\circ$.

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