A soccer ball is kicked off a bridge with a height of 36 m. The ball travels 25 m horizontally before it hits the pavement below. What was the soccer ball’s speed when it was first kicked?
3.7 m/s
To find the speed of the soccer ball when it was first kicked, we can use the principle of conservation of energy.
The initial potential energy of the ball when it was on top of the bridge is converted into kinetic energy as it falls.
The potential energy (PE) of the ball is given by the formula PE = mgh, where m is the mass of the ball (which we'll assume to be constant), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from which it is falling.
The kinetic energy (KE) of the ball is given by the formula KE = 0.5mv^2, where v is the velocity of the ball.
According to the conservation of energy, the initial potential energy is equal to the final kinetic energy. Therefore, we have:
mgh = 0.5mv^2
We can cancel out the mass from both sides of the equation, leaving us with:
gh = 0.5v^2
Now we can solve for v by substituting the given values:
g = 9.8 m/s^2
h = 36 m
Plugging these values into the equation:
(9.8 m/s^2)(36 m) = 0.5v^2
331.2 m^2/s^2 = 0.5v^2
Dividing both sides by 0.5:
662.4 m^2/s^2 = v^2
Taking the square root of both sides:
v ≈ 25.73 m/s
Therefore, the speed of the soccer ball when it was first kicked is approximately 25.73 m/s.