v = Vi - g t
0 = 25 - 10 t
t = 2.5 seconds to top
h = Vi t - 1/2 g t^2
h = 25 * 2.5 - 5 * 6.25
0 = 25 - 10 t
t = 2.5 seconds to top
h = Vi t - 1/2 g t^2
h = 25 * 2.5 - 5 * 6.25
h = (v^2 - u^2) / (2g)
where:
h = height
v = final velocity (0 m/s at the highest point)
u = initial velocity (25.0 m/s)
g = acceleration due to gravity (10 m/s^2)
First, let's find the height (h):
h = (0^2 - 25.0^2) / (2 * -10)
h = 625 / -20
h = -31.25 meters
The negative sign indicates that the ball falls back down to its starting point. Therefore, the ball reaches a maximum height of 31.25 meters above its starting point.
To find the time it takes to reach the highest point, we can use another kinematic equation:
v = u + gt
where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (25.0 m/s)
g = acceleration due to gravity (10 m/s^2)
t = time
0 = 25.0 + (-10)t
10t = 25.0
t = 2.5 seconds
So, it takes 2.5 seconds for the ball to reach its highest point.