Use a(t) = -9.8 meters per second per second as the acceleration due to gravity. (Neglect air resistance.)
A baseball is thrown upward from a height of 3 meters with an initial velocity of 7 meters per second. Determine its maximum height. (Round your answer to two decimal places.)
To determine the maximum height of the baseball, we need to find the highest point it reaches before it starts falling back down.
We know the acceleration due to gravity is -9.8 meters per second per second, which means the velocity of the baseball is decreasing at a rate of 9.8 meters per second every second.
The initial velocity of the baseball is 7 meters per second, and it is thrown upward, so the velocity is initially positive.
First, let's find the time it takes for the baseball to reach its peak. We can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
The final velocity at the highest point is 0 m/s because the baseball momentarily stops before falling back down. The initial velocity is 7 m/s, and the acceleration is -9.8 m/s^2.
0 = 7 + (-9.8)t
Solving for t:
-9.8t = -7
t = -7 / -9.8
t ≈ 0.71 seconds
Now that we know the time it takes for the baseball to reach its peak, we can calculate the maximum height it reaches using the equation:
s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
Since we are measuring the height from a reference point of 3 meters, the displacement is the difference between the maximum height and the initial height, which is (s - 3).
Substituting the known values:
(s - 3) = 7 * 0.71 + (1/2) * (-9.8) * (0.71)^2
(s - 3) ≈ 4.97 - 2.47
(s - 3) ≈ 2.50 meters
s ≈ 2.50 + 3
s ≈ 5.50 meters
Therefore, the maximum height the baseball reaches is approximately 5.50 meters.
v = vi - a t
0 = 7 - 9.8 t
t = 7/9.8 at top when v = 0
h = 3 + 7 t - 4.9 t^2