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?
Calculate the time it takes the ball to fall 3 meters when thrown horizontally.
That time, t, is given by the formula
3.0 = (g/2) t^2. Solve for t.
After you have calculated t, you can get the speed of the ball using
V = (15 m)/t.
To find the speed of the ball as it leaves the player's hand, we can use the kinematic equation for horizontal motion. The equation we will use is:
range = (initial velocity) × (time)
Here, the range is given as 15 m, and the initial vertical velocity (since the ball is thrown horizontally) is 0 m/s. We can rearrange the equation to solve for the initial velocity:
(initial velocity) = range / (time)
Now, we need to find the time it takes for the ball to travel 15 m horizontally. We can use the equation for vertical motion to find the time of flight. The equation we will use is:
vertical displacement = (initial velocity) × (time) + (acceleration due to gravity) × (time^2) / 2
The vertical displacement is given as 3.0 m, the initial velocity is 0 m/s, and the acceleration due to gravity is 9.8 m/s^2. Rearranging the equation, we get:
(time^2) - (6.0 / 9.8) × (time) = 0
Simplifying further, we get:
(time^2) - (0.6122 × time) = 0
Using the quadratic formula, we can solve for the time:
time = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -0.6122, and c = 0. Plugging in these values, we find:
time = (-(-0.6122) ± √((-0.6122)^2 - 4(1)(0))) / (2(1))
time ≈ (0.6122 ± √(0.37508)) / 2
time ≈ (0.6122 ± 0.6120) / 2
Therefore, the possible values for the time are:
time ≈ (0.6122 + 0.6120) / 2 ≈ 0.6121 s
time ≈ (0.6122 - 0.6120) / 2 ≈ 0.0001 s
Since the time of flight cannot be 0.0001 seconds, we disregard that value.
Substituting the time into the equation for the initial velocity, we find:
(initial velocity) = 15 m / 0.6121 s ≈ 24.48 m/s
Therefore, the speed of the ball as it leaves the player's hand is approximately 24.48 m/s.
To determine the speed of the ball as it leaves the player's hand, we can use the principles of projectile motion. The horizontal motion can be considered independent of the vertical motion. Since the ball is thrown horizontally, there is no initial vertical velocity.
First, let's calculate the time it takes for the ball to reach the ground. We can use the formula:
h = (1/2) * g * t^2,
where h is the vertical displacement (in this case, 3.0 m), g is the acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time.
Rearranging the formula, we have:
t = sqrt(2h / g).
Substituting the given values, we have:
t = sqrt(2 * 3.0 / 9.8).
t = sqrt(0.6122).
t ≈ 0.782 s.
Now, since the horizontal motion is independent, we can calculate the horizontal distance traveled by using the equation:
d = v * t,
where d is the horizontal distance (15 m), v is the horizontal velocity, and t is the time.
Rearranging the formula, we have:
v = d / t.
Substituting the given values, we have:
v = 15 / 0.782.
v ≈ 19.18 m/s.
Therefore, the speed of the ball as it leaves the player's hand is approximately 19.18 m/s.