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CGP EDU Academic Team
Published on: September 12, 2026
The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t - 5t^{2}) meter and \(\mathbf{x} = 6t\) meter, where t is in second. The velocity with which the projectile is projected is
Text Solution
Verified by ExpertsThe correct answer is:
C
\(v_y = - \frac{dy}{dt} - 8t - 10t\) , \(v_{r} = -\frac{d\alpha}{dt} - 6\)
at the time of projection i.e. \(v_y = -\frac{dy}{dt} = -8\) and \(\nu_{s} \quad 6\)
\(\therefore v = \sqrt{v_x^2 + v_y^2} = \sqrt{6^2 + 8^2} = 10 \text{ m/s}\)
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