Question

An object is observed to fall from a bridge, striking the water below with a velocity of 187.82 m/s. The height of the bridge in meters is:

Answer 1

The height of the bridge is 1796.895 m if an object falls from the bridge and strike the water below with a velocity of 187.82 m/sec.

Velocity of the object, v = 187.80 m/s

Height of the bridge will be

According to the three equations of motion,

1) First equation Velocity - Time relation states that,

$v = u + at$

here 'a' will be the acceleration due to gravity i.e g = 9.8 $\frac{m}{secx^{2} }$

'v' = 187.80  m/s

'u' = 0 m/s

therefore,

$t = \frac{v - u}{a}$

$t = \frac{187.80 - 0}{9.81}$

t = 19.14 sec

2) Second equation Position - Time relation states that,

$s = ut + \frac{1}{2} at^{2}$

here 'a' will be the acceleration due to gravity i.e g = 9.8 $\frac{m}{secx^{2} }$

'u' = 0 m/s

't' = 19.14 sec

$s = 0 + \frac{1}{2} 9.81 *(19.14^{2})$

s = 0 + 4.905 * 366.34

s = 1796.895 m

The height of the the bridge is 1796.895 m.

To know more about three equations of motion here

brainly.com/question/14278353

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