A baseball is thrown at an angle of 29 degrees relative to the ground at a speed of 24.3 m/s. The ball is caught 51.0463 m from the thrower.
The acceleration due to gravity is 9.81 m/s2
How high is the tallest spot in the ball's path?
Answer in units of m
Vo = 24.3 m/s.[29o]
Yo = 24.3*sin29 = 11.8 m/s.
Yf = Yo + g*Tr = 0 at max h.
11.8 - 9.81Tr = 0.
Tr = 1.20 s. = Rise time.
h = Yo*Tr + 0.5g*Tr^2.
g = -9.81 m/s^2.
To find the height of the tallest spot in the ball's path, we can first determine the time it takes for the ball to reach its highest point.
1. Start by breaking down the initial velocity of the ball into its x and y components. The vertical component can be found using the equation:
Vy = V * sin(theta), where V is the initial velocity (24.3 m/s) and theta is the angle of 29 degrees.
Vy = 24.3 m/s * sin(29 degrees)
Vy = 24.3 m/s * 0.483
Vy = 11.74 m/s
2. Next, we can use the equation for vertical motion to find the time it takes for the ball to reach its highest point:
Vy = Vo + at, where a is the acceleration due to gravity (-9.81 m/s^2) and Vo is the vertical component of the initial velocity.
0 m/s = 11.74 m/s + (-9.81 m/s^2) * t
9.81 m/s^2 * t = 11.74 m/s
t = 11.74 m/s / 9.81 m/s^2
t ≈ 1.195 s
3. Now that we have the time it takes for the ball to reach its highest point, we can find the maximum height using the equation for vertical displacement:
Δy = Vo * t + 0.5 * a * t^2, where Δy is the vertical displacement, Vo is the vertical component of the initial velocity, and a is the acceleration due to gravity.
Δy = 11.74 m/s * 1.195 s + 0.5 * (-9.81 m/s^2) * (1.195 s)^2
Δy ≈ 7.85 m
Therefore, the height of the tallest spot in the ball's path is approximately 7.85 meters.
To determine the height of the tallest spot in the ball's path, we need to analyze the ball's motion in the vertical direction.
First, let's break down the initial velocity of the ball into its vertical and horizontal components. We can use trigonometric functions to find the vertical component:
Initial vertical velocity (Vy) = Initial velocity (V) * sin(angle)
= 24.3 m/s * sin(29°)
= 11.66 m/s (rounded to two decimal places)
Next, we can determine the time it takes for the ball to reach its peak height. At the highest point of its trajectory, the vertical velocity will become zero. Using the kinematic equation:
Vy = V0y + at
Where:
Vy is the final vertical velocity (zero in this case),
V0y is the initial vertical velocity,
a is the acceleration due to gravity (-9.81 m/s^2),
and t is the time.
0 = 11.66 m/s - 9.81 m/s^2 * t
Solving for t:
t = 11.66 m/s / 9.81 m/s^2
≈ 1.19 s (rounded to two decimal places)
Now, let's find the vertical displacement (height) at the peak of the ball's path using the kinematic equation:
Δy = V0y * t + (1/2) * a * t^2
Where:
Δy is the vertical displacement (height),
V0y is the initial vertical velocity,
a is the acceleration due to gravity (-9.81 m/s^2),
and t is the time.
Δy = 11.66 m/s * 1.19 s + (1/2) * (-9.81 m/s^2) * (1.19 s)^2
≈ 6.94 m (rounded to two decimal places)
Therefore, the height of the tallest spot in the ball's path is approximately 6.94 meters.