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A boy standing on the bridge kicks a stone into the water below. He kicks the stone with a horizontal velocity of 8.06 m/s. It lands in the water 6.78 m away from the bridge. How high is the bridge?

Answer :

Time taken to reach water :

[tex]t = \dfrac{D}{v}\\\\t=\dfrac{6.78}{8.06}\ s\\\\t=0.84\ s[/tex]

Now, initial vertical speed , u = 0 m/s.

By equation of motion :

[tex]h = ut +\dfrac{at^2}{2}[/tex]

Here, a = g = acceleration due to gravity = 9.8 m/s².

So,

[tex]h = 0(t) +\dfrac{9.8\times 0.84^2}{2}\\\\h=3.46\ m[/tex]

Therefore, the height of the bridge is 3.46 m.

Hence, this is the required solution.

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