Question

A body is projected from a certain height with an initial velocity \(u\) making an angle \(\theta\) above the horizontal. The time taken for its vertical and horizontal displacements to become equal is:

• A. \(\frac{2u(\sin\theta-\cos\theta)}{g}\)
• B. \(\frac{u(\sin\theta+\cos\theta)}{g}\)
• C. \(\frac{u(\sin\theta-\cos\theta)}{g}\)
• D. \(\frac{2u(\sin\theta+\cos\theta)}{g}\)

Hint:

Whenever a projectile question involves equality of horizontal and vertical displacements, write \(x=u\cos\theta\,t\) and \(y=u\sin\theta\,t-\frac12 gt^2\), then equate them directly.

Answer 1

The Correct Option is A

Concept: For projectile motion, horizontal and vertical motions are treated independently. Horizontal displacement: \[ x=u\cos\theta\; t \] Vertical displacement: \[ y=u\sin\theta\; t-\frac12 gt^2 \] The condition given is that the horizontal and vertical displacements become equal.

Step 1: Write the condition of equality. Given \[ x=y \] Substituting the displacement equations, \[ u\cos\theta\, t = u\sin\theta\, t-\frac12 gt^2 \]

Step 2: Rearrange the equation. Bringing all terms to one side, \[ u\sin\theta\, t-u\cos\theta\, t = \frac12 gt^2 \] Factorizing \(t\), \[ ut(\sin\theta-\cos\theta) = \frac12 gt^2 \]

Step 3: Cancel common factor. For non-zero time, \[ u(\sin\theta-\cos\theta) = \frac12 gt \] Multiplying both sides by 2, \[ 2u(\sin\theta-\cos\theta) = gt \]

Step 4: Calculate the required time. Therefore, \[ t = \frac{2u(\sin\theta-\cos\theta)}{g} \]

Step 5: Final answer. \[ \boxed{\frac{2u(\sin\theta-\cos\theta)}{g}} \]