Question

For the projectile shown in figure what will be angle of projection?

• A. 53°
• B. $\tan^{-1}(\frac{7}{4})$
• C. $\sin^{-1}(\frac{2}{7})$
• D. $\tan^{-1}(\frac{4}{7})$

Answer 1

Answer: (B)

$\tan^{-1}(\frac{7}{4})$

Answer 2

To find the angle of projection ($\theta$), we can use the properties of projectile motion.

Let the projectile be launched from the origin (0,0) with an initial velocity $u$ at an angle $\theta$ with the horizontal. The projectile lands at a horizontal distance $R$ from the origin.

The equation of the trajectory of a projectile is given by: $y = x \tan\theta \left(1 - \frac{x}{R}\right)$

From the figure, let P(x, y) be the point on the trajectory. The angle formed by the line joining the origin (0,0) to P(x,y) with the horizontal is $\alpha = 37^\circ$. So, $\tan\alpha = \frac{y}{x}$. Given $\alpha = 37^\circ$, we know $\tan 37^\circ = \frac{3}{4}$. Therefore, $\frac{y}{x} = \frac{3}{4}$ (Equation 1)

The angle formed by the line joining the landing point (R,0) to P(x,y) with the horizontal is $\beta = 45^\circ$. So, $\tan\beta = \frac{y}{R-x}$. Given $\beta = 45^\circ$, we know $\tan 45^\circ = 1$. Therefore, $\frac{y}{R-x} = 1$ (Equation 2)

From Equation 1, $y = \frac{3}{4}x$. From Equation 2, $y = R-x$.

Equating the expressions for $y$: $\frac{3}{4}x = R-x$ $\frac{3}{4}x + x = R$ $\frac{7}{4}x = R$ $x = \frac{4}{7}R$

Now substitute the value of $x$ back into the expression for $y$: $y = \frac{3}{4}x = \frac{3}{4} \left(\frac{4}{7}R\right) = \frac{3}{7}R$

So, the coordinates of the point P are $\left(\frac{4}{7}R, \frac{3}{7}R\right)$.

Now, substitute these coordinates into the trajectory equation: $y = x \tan\theta \left(1 - \frac{x}{R}\right)$ $\frac{3}{7}R = \left(\frac{4}{7}R\right) \tan\theta \left(1 - \frac{\frac{4}{7}R}{R}\right)$ $\frac{3}{7}R = \frac{4}{7}R \tan\theta \left(1 - \frac{4}{7}\right)$ $\frac{3}{7}R = \frac{4}{7}R \tan\theta \left(\frac{3}{7}\right)$

Cancel $R$ from both sides (assuming $R \neq 0$): $\frac{3}{7} = \frac{4}{7} \tan\theta \left(\frac{3}{7}\right)$

Cancel $\frac{3}{7}$ from both sides (assuming $\frac{3}{7} \neq 0$): $1 = \frac{4}{7} \tan\theta$ $\tan\theta = \frac{7}{4}$

Therefore, the angle of projection is $\theta = \tan^{-1}\left(\frac{7}{4}\right)$.

Alternatively, a useful formula for this scenario is: If a projectile is launched from the origin with angle $\theta$ and range $R$, and a point P on its trajectory makes angles $\alpha$ and $\beta$ with the horizontal from the origin and the landing point respectively, then: $\tan\theta = \tan\alpha + \tan\beta$

In this problem, $\alpha = 37^\circ$ and $\beta = 45^\circ$. $\tan 37^\circ = \frac{3}{4}$ $\tan 45^\circ = 1$

Using the formula: $\tan\theta = \frac{3}{4} + 1 = \frac{3+4}{4} = \frac{7}{4}$ $\theta = \tan^{-1}\left(\frac{7}{4}\right)$

The angle of projection is $\tan^{-1}(\frac{7}{4})$. The correct option is (b).