You want to calculate the displacement of an object thrown over a bridge. Using -10 m/s2 for acceleration due to gravity, what would be the total displacement of the object if it took 8 seconds before hitting the water?
for your situation
distance = -5t^2 , if it was dropped. If it was thrown with some velocity up or down , we would have to know that
so distance = -5(8^2) = -320 m , negative because downwards
(tall bridge, considering that the tallest bridge in the US, The Royal Gorge bridge, is about 290 m)
To calculate the total displacement of the object, we need to take into account the initial velocity, acceleration due to gravity, and the time it takes for the object to hit the water.
In this case, the object is thrown over a bridge, so we can assume it has an initial velocity of zero (since it is not given). The acceleration due to gravity is -10 m/s^2, which is negative because it acts in the opposite direction of the object's motion.
Given that the object takes 8 seconds before hitting the water, we can use the formula for displacement:
displacement = initial velocity * time + (1/2) * acceleration * time^2
Since the initial velocity is zero, it simplifies to:
displacement = (1/2) * acceleration * time^2
Substituting the values:
displacement = (1/2) * (-10 m/s^2) * (8 s)^2
Simplifying the equation:
displacement = (1/2) * (-10 m/s^2) * 64 s^2
displacement = -320 m
Therefore, the total displacement of the object thrown over the bridge and hitting the water after 8 seconds would be -320 meters.