Question

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 25 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 21 m/s. How high is the bridge

Answer 1

Answer:

the bridge has a height y₀ = 6.94 m

Step-by-step explanation:

The position y of the loose bolt is given by (0,y) where

y = y₀ - 1/2*g*t²

where

y₀ = initial position of the bolt (height of the bridge) , g= gravity , t=time

and the position x of the car is given by (x,0) where

x= x₀  + v*t

where

x₀= initial position of the car

v= car's velocity

then in order for the bolt to hit the windshield they should be at  x=0 and y=0 at the same time , then

0= x₀  + v*t

t= -x₀/v

replacing in the equation for y

0 = y₀ - 1/2*g*t²

0 = y₀ - 1/2*g*(-x₀/v)²

0 = y₀ - 1/2*g*x₀²/v²

y₀ =  1/2*g*x₀²/v²

replacing values

y₀ =  1/2*g*x₀²/v² = 1/2* 9.8m/s² * (-25 m)²/(21 m/s)² = 6.94 m

then the bridge has a height y₀ =6.94 m

We have assumed that

- The bolt has no horizontal velocity ( only vertical velocity) , starts from rest and neglected air friction

- Neglecting the height of the car , position of the windshield and size of the loose bolt