Projectile Motion: Time to Reach Maximum Height

This question asks us to find the time it takes for a projectile, launched at an angle θ to the horizontal, to reach its highest point. We are given that the maximum height achieved is H.

Understanding Key Concepts

In projectile motion, the trajectory is influenced by gravity acting vertically downwards. The motion can be analyzed separately in horizontal (x) and vertical (y) directions.

• Initial Velocity Components: If the initial velocity is u, the horizontal component is $u_x = u \cos \theta$ and the vertical component is $u_y = u \sin \theta$.
• Motion at Maximum Height: At the maximum height (H), the vertical component of the projectile's velocity becomes zero ($v_y = 0$). The horizontal component ($v_x = u_x$) remains constant throughout the flight (ignoring air resistance).
• Time to Reach Maximum Height: We can use the kinematic equation for vertical motion: $v_y = u_y - gt$, where g is the acceleration due to gravity and t is the time.

Deriving the Time to Maximum Height

Let $t_{top}$ be the time taken to reach the maximum height. At the maximum height:

• $v_y = 0$
• $u_y = u \sin \theta$

Using the kinematic equation $v_y = u_y - gt_{top}$: $$0 = u \sin \theta - gt_{top}$$ Rearranging this equation to solve for $t_{top}$: $$gt_{top} = u \sin \theta$$ $$t_{top} = \frac{u \sin \theta}{g}$$

Relating Maximum Height (H) to Initial Velocity

The formula for the maximum height (H) attained by a projectile is given by:

We can rearrange this formula to find an expression for $u \sin \theta$ in terms of H and g:

Multiply both sides by $2g$: $$2gH = (u \sin \theta)^2$$ Take the square root of both sides: $$u \sin \theta = \sqrt{2gH}$$

Calculating the Final Time

Now, substitute the expression for $u \sin \theta$ back into the formula for $t_{top}$:

To simplify this expression, we can rewrite g as $\sqrt{g^2}$: $$t_{top} = \frac{\sqrt{2gH}}{\sqrt{g^2}}$$ Combine the terms under a single square root: $$t_{top} = \sqrt{\frac{2gH}{g^2}}$$ Simplify the fraction inside the square root: $$t_{top} = \sqrt{\frac{2H}{g}}$$

Conclusion

The time taken by the projectile to reach its highest point is $\sqrt {\frac{{2H}}{g}}$. This matches option 2.