To find how high the ball goes, we can use the kinematic equation that relates initial velocity (u), final velocity (v), acceleration (a), and displacement (s):
v^2 = u^2 + 2as
In this case, the ball is thrown vertically, so its initial velocity (u) is 25 m/s, and at the highest point of its trajectory, its final velocity (v) is 0 m/s. We need to find the displacement (s), which represents the height the ball reaches.
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Since the ball is thrown vertically, the only force acting on it is gravity, resulting in an acceleration (a) of approximately -9.8 m/s^2 (taking into account the negative sign to indicate it acts in the opposite direction to the upward motion).
Substituting the values into the equation, we get:
s = (0^2 - 25^2) / (2 * -9.8) s = (-625) / (-19.6) s ≈ 31.88 m
Therefore, the ball reaches a height of approximately 31.88 meters.
To find the velocity of the ball when it reaches a height of 25 meters, we can use another kinematic equation that relates initial velocity (u), final velocity (v), acceleration (a), and displacement (s):
v^2 = u^2 + 2as
This time, we want to find the final velocity (v) when the ball is at a height of 25 meters. We already know the initial velocity (u) is 25 m/s, and the acceleration (a) is approximately -9.8 m/s^2.
Rearranging the equation, we have:
v = sqrt(u^2 + 2as)
Substituting the values, we get:
v = sqrt(25^2 + 2 * (-9.8) * 25) v = sqrt(625 - 490) v = sqrt(135) v ≈ 11.62 m/s
Therefore, the velocity of the ball when it reaches a height of 25 meters is approximately 11.62 m/s.