The velocity of a particle at t=0 is u→=(4i^+4j^)ms−1and its acceleration is −(10j^)ms−2. At t = 0, the particle is at (1, 0) m. The x-coordinate of the point, where its y-coordinate is again zero is (in m).

Here, x0=1m, y0=0ux=4ms−1,uy=4ms−1ax=0 and ay=−10ms−2∵y−y0=uyt+12ayt2or y−0=4t−12×10t2or 0=4t−5t2∴ t=45s∵ x−x0=uxtor x−1=4×45∴x=165+1=215m=4.2m

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The velocity of a particle at $t = 0$ is $\vec{u} = (4 \hat{i} + 4 \hat{j}) {ms}^{- 1}$ and its acceleration is $- (10 \hat{j}) {ms}^{- 2} .$ At t = 0, the particle is at (1, 0) m. The x-coordinate of the point, where its y-coordinate is again zero is (in m).

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answer is 4.2.

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Detailed Solution

Here, $x_{0} = 1 m, y_{0} = 0$

$\begin{array}{l} u_{x} = 4 {ms}^{- 1}, u_{y} = 4 {ms}^{- 1} \\ a_{x} = 0 and a_{y} = - 10 {ms}^{- 2} \\ ∵ y - y_{0} = u_{y} t + \frac{1}{2} a_{y} t^{2} \\ or y - 0 = 4 t - \frac{1}{2} \times 10 t^{2} \\ or 0 = 4 t - 5 t^{2} \\ ∴ t = \frac{4}{5} s \\ ∵ x - x_{0} = u_{x} t \\ or x - 1 = 4 \times \frac{4}{5} \\ ∴ x = \frac{16}{5} + 1 = \frac{21}{5} m = 4 .2 m \end{array}$