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# Question 41

A pebble is thrown vertically upwards with a speed of 20m/s. How high will it be after 2s? (Take g= 10m/s^2).

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Given: initial velocity u = 20 m/s, t=2s, g= 10m/s^2

The greatest height reached by the pebble can be obtained by the equation of motion

\begin{array}{l} \mathrm{H}=\mathrm{ut}+1 / 2 \mathrm{gt}^{2} \\ =0(2)+1 / 2(10)(2)^{2} \\ =20 \mathrm{m} \end{array}

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