a pitches takes 0.10 seconds to throw a baseball, which leaves his hand at a velocity of 30m/s the ball's acceleration is?
While being thrown (but not afterwards), the acceleration is
a = V/t = 300 m/s^2
bbb
To find the ball's acceleration, we can use the formula:
acceleration (a) = (final velocity (v) - initial velocity (u)) / time (t)
Given:
- Initial velocity (u) = 0 m/s (as the ball leaves the hand with a velocity of 30 m/s, assuming it starts from rest)
- Final velocity (v) = 30 m/s
- Time (t) = 0.10 seconds
Let's substitute the values into the formula to calculate the acceleration:
acceleration (a) = (30 m/s - 0 m/s) / 0.10 seconds
acceleration (a) = 30 m/s / 0.10 s
a = 300 m/s²
Therefore, the ball's acceleration is 300 m/s².
To find the ball's acceleration, we need to use the equation for acceleration: acceleration (a) = change in velocity (Δv) / time taken (Δt).
In this case, we know the initial velocity (v0) is 0 m/s because the ball starts from rest in the pitcher's hand. The final velocity (v) is given as 30 m/s. The time taken (Δt) is 0.10 seconds.
To calculate the change in velocity (Δv), we subtract the initial velocity from the final velocity: Δv = v - v0 = 30 m/s - 0 m/s = 30 m/s.
Now, substitute the values into the acceleration equation:
acceleration (a) = Δv / Δt = 30 m/s / 0.10 s.
Simply dividing 30 m/s by 0.10 s, we find the ball's acceleration:
a = 300 m/s².
Therefore, the ball's acceleration when it leaves the pitcher's hand is 300 m/s².