Baseball of 0.15 kg is thrown upward at an initial velocity of 35 m/s. tow second later a 20 g bullet is fired at 250 m/s into the rising baseball . how high will the combined bullet and baseball rise.
To solve this problem, we need to analyze the motion of both the baseball and the bullet independently and then determine the combined height they will reach.
First, let's consider the motion of the baseball. We know its mass (0.15 kg) and initial velocity (35 m/s).
Using the equations of motion, we can calculate the maximum height that the baseball will reach. The equation we will use is:
v^2 = u^2 + 2as
Where:
v = final velocity (which is 0 m/s at the maximum height)
u = initial velocity (35 m/s)
a = acceleration (which is -9.8 m/s^2 due to gravity)
s = distance or displacement
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Plugging in the values:
s = (0^2 - 35^2) / (2 * -9.8)
s = 0 - 1225 / -19.6
s = 62.5 meters
So, the baseball alone will reach a height of 62.5 meters.
Next, let's consider the motion of the bullet. We know its mass (20 g = 0.02 kg) and initial velocity (250 m/s).
Again, using the same equations of motion, we can calculate the maximum height that the bullet will reach:
s = (v^2 - u^2) / (2a)
Plugging in the values:
s = (0^2 - 250^2) / (2 * -9.8)
s = 0 - 62500 / -19.6
s = 3198.98 meters
So, the bullet alone will reach a height of approximately 3198.98 meters.
Now, to find the combined height, we add the individual heights achieved by the baseball and the bullet:
Combined height = 62.5 meters (baseball) + 3198.98 meters (bullet)
Combined height = 3261.48 meters
Therefore, the combined bullet and baseball will rise to a height of approximately 3261.48 meters.