A baseball is hit at 29.0 m/s at an angle of 50.0° with the horizontal. Immediately an outfielder runs 4.90 m/s toward the infield and catches the ball at the same height it was hit. What was the original distance between the batter and the outfielder?
A body of mass 2kg moving with a velocity of 6m/s collite with a stationary object of mass 500gm if the two bodies move together after Impact. Calculate their common velocity
To find the original distance between the batter and the outfielder, we can start by breaking down the initial velocity of the baseball into its horizontal and vertical components.
Given:
Initial velocity of the baseball (v) = 29.0 m/s
Angle with the horizontal (θ) = 50.0°
Outfielder's velocity (v_outfielder) = 4.90 m/s
First, we need to find the vertical component of the baseball's initial velocity. We can use trigonometry to determine this:
Vertical component of velocity (v_vertical) = v * sin(θ)
Next, we need to find the time it takes for the outfielder to catch the ball. Since we know the height at which the ball was hit and caught is the same, we can ignore the vertical motion and focus only on the horizontal motion:
Distance traveled horizontally (d_horizontal) = v * cos(θ) * t
Where t represents the time taken.
Now, to find the time (t), we can use the equation for constant velocity motion:
Distance traveled by the outfielder (d_outfielder) = v_outfielder * t
Since the outfielder catches the ball, d_outfielder = d_horizontal. By substituting the values we have, we can solve for t.
Once we have the value of t, we can substitute it back into the equation for d_horizontal to find the original distance between the batter and the outfielder.