Question

The equation of the trajectory of a ball projected at an angle \(θ\) with the horizontal, is given as \(y = x-\frac{gx^2}{2}\)
The initial velocity of the ball is
[Given : \(tan45^{\circ} = 1\), \(cos45^{\circ} = \frac{1}{\sqrt{2}}\) ]

• A. \(2\sqrt{2}\,\text{m/s}\)
• B. \(2\,\text{m/s}\)
• C. \(\sqrt{2}\,\text{m/s}\)
• D. \(\frac{1}{\sqrt{2}}\,\text{m/s}\)

Hint:

Compare the given trajectory with y = x tan(theta) - g x^2 / (2 u^2 cos^2(theta)).

Answer 1

The Correct Option is C

Step 1: Understand the concept
A projectile launched with speed \(u\) at angle \(\theta\) follows
\[ y = x\tan\theta - \frac{g\,x^2}{2u^2\cos^2\theta} \]

Step 2: Compare with the given equation
The given equation is \(y = x - \frac{g x^2}{2}\). Matching the coefficient of \(x\): \(\tan\theta = 1\), so \(\theta = 45^\circ\).

Step 3: Match the \(x^2\) term
\[ \frac{g}{2u^2\cos^2\theta} = \frac{g}{2} \Rightarrow u^2\cos^2\theta = 1 \]
With \(\cos^2 45^\circ = \frac{1}{2}\), we get \(u^2 = 2\).

Step 4: Result
\(u = \sqrt{2}\) m/s, option (C). Option (A) would give \(u^2 = 8\) and (B) would give \(u^2 = 4\), which do not satisfy \(u^2\cos^2\theta = 1\).

Final Answer:
The initial speed is sqrt 2 m/s. This is option (C). \[ \boxed{\text{(C) }\sqrt{2}\ \text{m/s}} \]