A ball is thrown up and attains a maximum height of 1000 m.It's initial speed was?
V^2 = Vo^2 + 2g*h = 0
Vo^2 = -2g*h = -2*(-9.8)*1000 = 19,600
Vo = 140 m/s.
To calculate the initial speed (also known as the initial velocity or the launch speed) of the ball, we can use the kinematic equation for vertical motion:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (which is 0 when the ball reaches its maximum height),
- vi is the initial velocity,
- a is the acceleration due to gravity (approximately -9.8 m/s^2, assuming the ball is thrown on Earth),
- and d is the displacement (which is the maximum height of 1000 m).
Substituting these values into the equation, we get:
0^2 = vi^2 + 2(-9.8)(1000)
Simplifying further:
0 = vi^2 - 19600
Rearranging the equation:
vi^2 = 19600
Taking the square root of both sides to solve for vi:
vi = √19600
Therefore, the initial speed of the ball is approximately 140 m/s.