A ball player wishes to determine her pitching speed by throwing a ball horizontally from an elevation of 3.0 m above the ground. She sees the ball land 15 m down the range. What is the speed of the ball as it leaves her hand?
Duplicate post. Already answered.
To determine the speed of the ball as it leaves the player's hand, we can use the concept of projectile motion. When the ball is thrown horizontally, its initial horizontal velocity remains constant throughout its trajectory. The vertical motion, however, is affected by gravity.
First, we need to find the time it takes for the ball to hit the ground. We can use the equation:
h = u*t + (1/2)*g*t^2
Where:
h = vertical displacement (in this case, 3.0 m above the ground)
u = initial vertical velocity (0 m/s since the ball is thrown horizontally)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Rearranging the equation and substituting the given values, we have:
3.0 = 0*t + (1/2)*(-9.8)*t^2
Simplifying further:
1.5 = -4.9*t^2
Dividing both sides by -4.9, we get:
t^2 = -1.5/-4.9
t^2 ≈ 0.3061
Taking the square root of both sides, we find:
t ≈ √0.3061
t ≈ 0.553 s
Now that we have the time, we can find the horizontal distance covered by the ball. We can use the equation:
d = v*t
Where:
d = horizontal distance (15 m)
v = horizontal velocity (which is constant)
Rearranging the equation, we find:
v = d/t
Substituting the given values:
v = 15/0.553
v ≈ 27.1 m/s
Therefore, the speed of the ball as it leaves the player's hand is approximately 27.1 m/s.