A baseball is thrown at an angle of 26 degrees relative to the ground at a speed of 22.1 m/s. The ball is caught 39.2327 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 = 22.1m/s @ 26deg.
Yo = ver. = 22.1sin26 = 9.69m/s.
hmax = (Yf^2 - Yo^2) / 2g,
hmax = (0 - (9.69)^2) / -19.6 = 4.79m.
To find the height of the tallest spot in the ball's path, we need to analyze the motion of the ball in the vertical direction.
1. Determine the initial vertical velocity (Vy):
The vertical component of the initial velocity can be calculated using the equation Vy = V * sin(θ), where V is the initial velocity of the ball and θ is the angle of projection.
Plugging in the values, Vy = 22.1 m/s * sin(26°) = 9.91 m/s.
2. Determine the time taken to reach the highest point (t_h):
The time taken to reach the highest point can be found using the equation Vy = V0y + g * t, where g is the acceleration due to gravity.
At the highest point, the vertical component of the velocity (Vy) becomes zero. So, Vy = 0.
0 = 9.91 m/s + (-9.81 m/s^2) * t_h.
Solving for t_h, we get t_h = 1.01 s.
3. Determine the height of the highest point (h_h):
The height of the highest point can be calculated using the equation h_h = V0y * t_h + (1/2) * g * t_h^2, where V0y is the initial vertical velocity and g is the acceleration due to gravity.
Plugging in the values, h_h = 9.91 m/s * 1.01 s + (1/2) * (-9.81 m/s^2) * (1.01 s)^2.
Solving for h_h, we get h_h = 5.04 m.
Therefore, the highest spot in the ball's path is at a height of 5.04 meters.
To find the highest point of the ball's path, we can use the kinematic equations for projectile motion. The vertical motion of the ball can be treated separately from the horizontal motion.
First, let's find the time it takes for the ball to reach its highest point. We can use the equation:
Δy = v0y * t - 0.5 * g * t^2
where Δy is the vertical displacement, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
At the highest point, the vertical velocity is zero, so we can set v0y = 0.
0 = 0 - 0.5 * g * t^2
Simplifying the equation:
0.5 * g * t^2 = 0
Since g is positive and cannot be zero, t must be zero.
This means that the ball reaches its highest point instantly.
Now, let's find the vertical displacement, or the height, of the highest point. We can use the equation:
Δy = v0y * t - 0.5 * g * t^2
Since t = 0, the first term (v0y * t) becomes zero, leaving us with:
Δy = - 0.5 * g * t^2
However, since we want the magnitude or absolute value of the height, we can drop the negative sign.
Now plug in the values:
Δy = 0.5 * (9.81 m/s^2) * (0 s)^2
Δy = 0 m
Therefore, the highest point in the ball's path is at a height of 0 meters.