Question

A projectile is thrown at an angle 37° from the vertical. The angle of elevation of the highest point of the projectile from point of projection is

• A. tan$^{-1}$($\frac{3}{2}$)
• B. tan$^{-1}$($\frac{2}{3}$)
• C. tan$^{-1}$($\frac{3}{8}$)
• D. tan$^{-1}$($\frac{8}{3}$)

Answer 1

Answer: (B)

tan$^{-1}$($\frac{2}{3}$)

Answer 2

1. Determine the angle of projection from the horizontal:
   The projectile is thrown at an angle of 37° from the vertical. Therefore, the angle of projection from the horizontal, $\theta$, is:
   $\theta = 90^\circ - 37^\circ = 53^\circ$

2. Recall formulas for maximum height and horizontal range:
   Let $u$ be the initial velocity of the projectile.
   The maximum height ($H$) reached by the projectile is given by:
   $H = \frac{u^2 \sin^2 \theta}{2g}$
   The horizontal range ($R$) of the projectile is given by:
   $R = \frac{u^2 \sin(2\theta)}{g}$

3. Define the angle of elevation of the highest point:
   The highest point of the projectile's trajectory is at a horizontal distance of $R/2$ from the point of projection and at a vertical height of $H$.
   Let $\alpha$ be the angle of elevation of the highest point from the point of projection.
   From the geometry, we can write:
   $\tan \alpha = \frac{\text{Vertical height}}{\text{Horizontal distance to highest point}} = \frac{H}{R/2} = \frac{2H}{R}$

4. Substitute the formulas for H and R into the expression for $\tan \alpha$:
   $\tan \alpha = \frac{2 \left(\frac{u^2 \sin^2 \theta}{2g}\right)}{\left(\frac{u^2 \sin(2\theta)}{g}\right)}$
   $\tan \alpha = \frac{\frac{u^2 \sin^2 \theta}{g}}{\frac{u^2 \sin(2\theta)}{g}}$
   $\tan \alpha = \frac{\sin^2 \theta}{\sin(2\theta)}$

5. Simplify the expression using trigonometric identities:
   Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$:
   $\tan \alpha = \frac{\sin^2 \theta}{2 \sin \theta \cos \theta}$
   $\tan \alpha = \frac{\sin \theta}{2 \cos \theta}$
   $\tan \alpha = \frac{1}{2} \tan \theta$

6. Substitute the value of $\theta$:
   We found $\theta = 53^\circ$.
   We know that $\tan(53^\circ) = \frac{4}{3}$ (from a 3-4-5 right-angled triangle where 53° is opposite the side of length 4).
   $\tan \alpha = \frac{1}{2} \times \frac{4}{3}$
   $\tan \alpha = \frac{4}{6}$
   $\tan \alpha = \frac{2}{3}$

7. Find the angle $\alpha$:
   $\alpha = \tan^{-1}\left(\frac{2}{3}\right)$