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Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceeds to fall straight down and ends up breaking the windshield of a car passing under the bridge. The car was 27 m away from the point of impact when the bolt began to fall down; unfortunately, the driver did not notice it and proceeded at constant speed of 17 m/s. How high is the bridge? Or more precisely, how high are the railroad tracks above the windshield height? The acceleration due to gravity is 9.8 m/s 2 . Answer in units of m

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fabianb4235

Answer:

12.35m

Explanation:

Hello! To solve this problem we must consider the following:

1. The car moves with constant speed, which means that the distance traveled is equal to the multiplication of time by speed.

X = VT

we solve the equation for time

[tex]T=\frac{x}{V} =\frac{27}{17} =1.588s[/tex]

2. The bolt moves with constant acceleration, with acceleration of 9.81m / s ^ 2, so we could apply the following equation.

note=remember that "a uniformly accelerated motion", means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

[tex]Y= VoT-\frac{1}{2}gt^{2}[/tex]

where

Vo = Initial speed

=0

T = time

=1.588s

g =gravity=9.8m/s^2

Y =  bridge height

solving

[tex]Y= \frac{1}{2}gt^{2}\\Y=(0.5)(9.8)(1.588^2)=12.35m[/tex]

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