Respuesta :
The stone will take 2.89 seconds to hit the water.
The time required by the stone to hit the water is calculated by the second equaiton of motion
s=ut+[tex] \frac{1}{2} [/tex]gt^2
41=0×t+[tex] \frac{1}{2} [/tex]×9.81×t²
t=2.89 seconds
The time taken by the stone to reach the surface of the water from the bridge is [tex]\boxed{2.89\,{\text{s}}}[/tex].
Further Explanation:
Given:
The height of the bridge from the surface of the river is [tex]41\,{\text{m}}[/tex].
Concept:
As the stone is dropped from the bridge, the stone falls freely under the action of the acceleration due to gravity of the Earth. The stone follows the second equation of motion during its fall from the bridge.
The expression for the second equation of motion for a body falling freely under the gravity is written as.
[tex]s = {v_i}t + \dfrac{1}{2}g{t^2}[/tex]
Here, [tex]s[/tex] is the height covered by the stone, [tex]{v_i}[/tex] is the initial speed of the stone, [tex]g[/tex] is the acceleration due to gravity and [tex]t[/tex] is the time taken.
Since the stone is dropped from the bridge, the initial speed of the stone will be zero and the acceleration due to gravity of the Earth is [tex]9.8\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}[/tex].
Substitute the values in above expression.
[tex]\begin{aligned}41&= \left( {0 \times t} \right) + \left( {\frac{1}{2} \times 9.8 \times {t^2}} \right)\\t&= \sqrt {\frac{{82}}{{9.8}}} \,{\text{s}}\\&\approx 2.89\,{\text{s}}\\\end{aligned}[/tex]
Thus, the time taken by the stone to reach the surface of the water from the bridge is [tex]\boxed{2.89\,{\text{s}}}[/tex].
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Answer Details:
Grade: High School
Chapter: Motion in one dimension
Subject: Physics
Keywords: Bridge, stone, falls, water, acceleration due to gravity, drops, measures the time, hit the water, air resistance, height, 41 m.
