The equation of the trajectory of a ball projected at an angle θ with the horizontal, is given as y = x - gx²2 The initial velocity of the ball is [Given: tan 45° = 1, cos 45° = 1/√(2) ] (a) 2√(2) m/s (b) 2 m/s (c) √(2) m/s (d) 1/√(2) m/s

Answer: (c). Compare with y = xtan θ - (gx²)/(2u²cos ²θ): tan θ = 1 ⇒ θ = 45° (g)/(2u²cos ²θ) = g2 ⇒ u²cos ²θ = 1 ⇒ u² · 12 = 1 u = √2 m/s So the answer is (c) √2 m/s.

The equation of the trajectory of a ball projected at an angle θ with… | MHT CET 2026 (19 Apr, Shift 2)

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Projectile Basics & Trajectory

The equation of the trajectory of a ball projected at an angle $\theta$ with the horizontal, is given as $y = x - \frac{gx^{2}}{2}$

The initial velocity of the ball is

[Given: $\tan 45^\circ = 1,\ \cos 45^\circ = \frac{1}{\sqrt{2}}$ ]

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Answer: (c)

Compare with $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$ : $\tan\theta = 1 \Rightarrow \theta = 45^\circ$
$\frac{g}{2u^2\cos^2\theta} = \frac g2 \Rightarrow u^2\cos^2\theta = 1 \Rightarrow u^2 \cdot \frac12 = 1$
$u = \sqrt2$ m/s
So the answer is (c) $\sqrt2$ m/s.

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Projectile Basics & Trajectory question from MHT CET 2026 (19 Apr, Shift 2)

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