Question:

A stone is thrown with an initial speed of 4.9 m/s4.9\, m / s from a bridge in vertically upward direction. It falls down in water after 2 sec⁡2\, \sec. The height of bridge is

Updated On: Jun 7, 2022
  • 24.7 m
  • 19.8 m
  • 9.8 m
  • 4.9 m
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The Correct Option is C

Solution and Explanation

Here: Initial velocity u=−4.9 m/su=-4.9\, m / s (-ve is due to vertically upward motion)
Total time t=2 sec⁡t=2\, \sec
The height of the bridge is given by
s=ut+12gt2s =u t+\frac{1}{2} g t^{2}
=−4.9×2+12×9.8×(2)2=-4.9 \times 2+\frac{1}{2} \times 9.8 \times(2)^{2}
=−9.8+19.6=9.8 m=-9.8+19.6=9.8\, m
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Types of Linear Motion:

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