A ball is projected vertical upwards from a balcony 45 m about the ground at a velocity of 10 m/s and hits the ground eventually ignore the effects of friction and assume the ball moves in the same vertical plane.
1)Calculate the maximum height reached by Bo the ball above the balcony
To find the maximum height reached by the ball, we can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2aΔy
where:
vf = final velocity (0 m/s at the maximum height, since the ball momentarily stops)
vi = initial velocity (10 m/s)
a = acceleration (acceleration due to gravity, -9.8 m/s^2 since the ball moves upwards)
Δy = change in height (the maximum height we're trying to find)
Plugging in the values:
0^2 = (10 m/s)^2 + 2(-9.8 m/s^2)Δy
Simplifying:
0 = 100 m^2/s^2 - 19.6 m/s^2 Δy
19.6 m/s^2 Δy = 100 m^2/s^2
Δy = 100 m^2/s^2 / 19.6 m/s^2
Δy ≈ 5.1 m
Therefore, the maximum height reached by the ball above the balcony is approximately 5.1 meters.