A projectile is given an initial velocity of (i + 2j) m/s , where i is along the ground and j is along the vertical. If g = 10 m/s^2 , the equation of - Physics MCQ JEE Main 2026 | ExamTest.live

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A projectile is given an initial velocity of (\hat{i} + 2 \hat{j}) m/s, where \hat{i} is along the ground and \hat{j} is along the vertical. If g = 10 m/s^{2}, the equation of its trajectory is:

Solution & Step-by-step Explanation

1. Initial velocity u_{x} = 1, u_{y} = 2 .2. Displacement at time t : x = u_{x} t = t y = u_{y} t - \frac{1}{2} g t^{2} = 2 t - \frac{1}{2} (10) t^{2} = 2 t - 5 t^{2} 3. Substitute t = x into the y equation: y = 2 x - 5 x^{2} .

Try it yourself before checking the explanation above.

A projectile is given an initial velocity of (\hat{i} + 2 \hat{j}) m/s, where \hat{i} is along the ground and \hat{j} is along the vertical. If g = 10 m/s^{2}, the equation of its trajectory is:

y = 2x - 5x^2

y = x - 5x^2

4y = 2x - 25x^2

y = x - 2.5x^2

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