A ball is thrown straight up with an initial
speed of 30 m/s.
How high does it go? Assume the acceleration of gravity is 10 m/s
2
.
Answer in units of m
80uy
To determine how high the ball goes, we can use the following kinematic equation:
(vf)^2 = (vi)^2 + 2aΔy
Where:
vf is the final velocity (which is 0 since the ball reaches its maximum height),
vi is the initial velocity (30 m/s),
a is the acceleration (negative 10 m/s^2 because gravity opposes the ball's motion),
Δy is the displacement or height we want to find.
Rearranging the equation, we have:
2aΔy = -(vi)^2
Substituting the given values, we get:
2(-10 m/s^2)Δy = -(30 m/s)^2
Dividing both sides by -20 m/s^2, we get:
Δy = 45 m
Therefore, the ball goes to a height of 45 meters.