A baseball pitcher throws a pitch with an initial velocity of 39.9 m/s, directed horizontally. How far does the ball drop vertically by the time it crosses the plate 18.0 m away?
To calculate the vertical drop of the baseball by the time it crosses the plate, we need to consider the time it takes for the ball to cover the horizontal distance of 18.0 m.
The initial velocity of the baseball in the horizontal direction does not affect the vertical motion of the ball. Hence, we can ignore it for this calculation.
First, we need to find the time it takes for the ball to travel 18.0 m horizontally. We can use the formula:
distance = velocity x time
Rearranging the formula gives:
time = distance / velocity
Plugging in the values, we have:
time = 18.0 m / 39.9 m/s
Calculating this, we find that the time taken by the ball to cross the plate is approximately 0.451 s.
Now, to calculate the vertical drop, we can use the formula for vertical displacement under constant acceleration:
displacement = (1/2) x acceleration x time^2
The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Substituting the values, we have:
displacement = (1/2) x 9.8 m/s^2 x (0.451 s)^2
Calculating this, we find that the ball drops vertically approximately 0.994 m by the time it crosses the plate.
Therefore, the ball drops vertically by about 0.994 meters.