AP EAMCET · PHYSICS · Motion In Two Dimensions
At any instant ' \(\mathrm{t}\) ' the vertical distance ' \(\mathrm{Y}\) ' and horizontal distance ' \(\mathrm{X}\) ' of a projectile are given by \(2 \mathrm{Y}=6 \mathrm{t}-\mathrm{gt}^2\) and \(X=4 \mathrm{t}\). The initial velocity of projectile is ( \(\mathrm{X}\) and \(\mathrm{Y}\) are in metre and \(\mathrm{t}\) is in second)
- A \(3 \mathrm{~ms}^{-1}\)
- B \(4 \mathrm{~ms}^{-1}\)
- C \(5 \mathrm{~ms}^{-1}\)
- D \(6 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(5 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & x=4 t \\ & V_x=\frac{d x}{d t}=4 \\ & Y=3 t-\frac{g t^2}{2} \\ & V_y=\frac{d y}{d t}=3-g t \\ & \text { at } t=0 V_y=3\end{aligned}\) \(\begin{aligned} & V_i=\sqrt{V_x^2+V_y^2} \\ & =\sqrt{4^2+3^2}=5 \mathrm{~m} / \mathrm{s}\end{aligned}\)
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