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: Use components
Let $u_x = u\cos\theta$ and $u_y = u\sin\theta$. The trajectory is $y = \frac{u_y}{u_x}x - \frac{g}{2u_x^2}x^2$.

Step 2: Read off
From $y = x - \frac{g}{2}x^2$: $\frac{u_y}{u_x} = 1$ and $\frac{1}{u_x^2} = 1$. So $u_x = 1$ m/s and $u_y = 1$ m/s.

Step 3: Combine
$u = \sqrt{u_x^2 + u_y^2} = \sqrt{1 + 1} = \sqrt{2}$ m/s.

Step 4: Check
The projection angle has $\tan\theta = 1$, so $45^\circ$, in line with the given values of $\tan45^\circ$ and $\cos45^\circ$.

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