A baseball is thrown upward with an initial speed of 35 m/s. What is the speed at t = 2 s?
To determine the speed of the baseball at t = 2 s, we can use the equations of motion for an object under constant acceleration.
The equation we will use is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the baseball is thrown upward, so we can take the acceleration due to gravity as -9.8 m/s^2.
First, we need to find the initial velocity (u). The problem states that the baseball is thrown upward with an initial speed of 35 m/s, which means the initial velocity (u) is 35 m/s.
Next, we need to find the acceleration (a). Since the baseball is thrown upward, the acceleration is equal to the acceleration due to gravity, which is -9.8 m/s^2 (negative because it is acting in the opposite direction of the initial velocity).
Now we can plug in the values into the equation.
v = u + at
v = 35 m/s + (-9.8 m/s^2)(2 s)
Multiplying -9.8 m/s^2 by 2 s, we get:
v = 35 m/s - 19.6 m/s
v = 15.4 m/s
Therefore, the speed of the baseball at t = 2 s is 15.4 m/s.
v^2 - u^2 = 2as
v^2 = 2(9.8)(2) + 35^2
v^2 = 39.2 + 1225
v^2 = 1264.2
v = 35.5