To determine how high the ball goes when a girl throws it upwards with a velocity of 10 m/s, we can use the equations of motion.
The motion of the ball can be divided into two phases: the upward motion and the downward motion. At the highest point of its trajectory, the ball momentarily comes to rest before starting to descend.
To calculate the height reached by the ball, we need to find the maximum height it reaches during its upward motion. The equation that relates the final velocity (Vf), initial velocity (Vi), acceleration due to gravity (g), and height (h) is:
Vf^2 = Vi^2 + 2gh
At the highest point, the final velocity is zero (Vf = 0), and the initial velocity is 10 m/s (Vi = 10 m/s). The acceleration due to gravity is approximately 9.8 m/s^2. Substituting these values into the equation, we can solve for the height (h):
0^2 = (10 m/s)^2 + 2 * 9.8 m/s^2 * h
0 = 100 m^2/s^2 + 19.6 m/s^2 * h
-100 m^2/s^2 = 19.6 m/s^2 * h
h = -100 m^2/s^2 / 19.6 m/s^2
h ≈ -5.1 meters
The negative sign indicates that the height is measured relative to the initial position of the ball. Therefore, the ball reaches a maximum height of approximately 5.1 meters above its starting point.