a coin is thrown horizontally with a speed of 12m/s from the top of a bridge . if the coin strikes the sea water below the bridge after 3 find the height of the bridge above sea level
s = 1/2 at^2 = 4.9*9 = ...
The horizontal speed of the coin makes no difference in how fast it falls.
To determine the height of the bridge above sea level, we can start by analyzing the motion of the coin.
Assuming there are no external forces acting on the coin (such as air resistance), the vertical motion of the coin can be treated as free fall motion. We can use the equation:
h = vi * t + (1/2) * g * t^2
where:
h = height of the bridge above sea level
vi = initial vertical velocity of the coin (0 m/s since the coin is thrown horizontally)
t = time taken for the coin to hit the sea water (3 s, as given in the question)
g = acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the known values into the equation, we get:
h = 0 * 3 + (1/2) * 9.8 * 3^2
h = 0 + (1/2) * 9.8 * 9
h = 0 + 4.9 * 9
h = 44.1 meters
Therefore, the height of the bridge above sea level is 44.1 meters.