A ball is thrown from a point 1.0 m above the ground. The initial velocity is 19.6 m/s at an angle of 30 degrees above the horizontal. Find the maximum height of the ball above the ground. Calculate the speed of the ball at the highest point in the trajectory.

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max height is

h = v^2 sin^2(θ)/2g
= 19.6^2 * sin^2(30°)/(2*9.8)
= 4.9m

the speed at max height is just the horizontal component, so

19.6cos(30°) = 16.97m/s

Well, well, well, looks like we have a ball on the loose! Let's get to the bottom of this, shall we?

To find the maximum height of the ball above the ground, we need to analyze its trajectory. When the ball reaches the highest point of its trajectory, its vertical velocity becomes zero. So, we need to find the time it takes for the ball to reach that point.

We can start by breaking down the initial velocity of the ball into its vertical and horizontal components. The vertical component would be 19.6 m/s * sin(30°), and the horizontal component would be 19.6 m/s * cos(30°).

Now, the time it takes for the ball to reach the highest point can be found using the formula: t = (Vf - Vi) / a, where Vf is the final velocity, Vi is the initial velocity, and a is the acceleration (in this case, the acceleration due to gravity, which is approximately 9.8 m/s^2).

Since the vertical velocity at the highest point is zero, the final velocity (Vf) in the vertical direction would be zero. Therefore, we can calculate the time it takes for the ball to reach the highest point.

After finding the time, we can substitute it back into the equation s = Vit + 0.5at^2, where s is the displacement, Vi is the initial velocity, t is the time, and a is the acceleration. By using this equation, we can calculate the maximum height of the ball above the ground.

Now, since we've determined the maximum height, let's move on to calculating the speed of the ball at that point. Since speed is the magnitude of velocity, we can find it using the Pythagorean theorem.

The speed (v) at any given point can be calculated using the equation v = sqrt(vx^2 + vy^2), where vx is the horizontal component of velocity, and vy is the vertical component of velocity.

So, we can find the speed of the ball at the highest point by calculating vx and vy using the initial velocity and then plugging them into the equation.

Now, let's get calculating and track down that ball's height and speed!

To find the maximum height of the ball above the ground, we can use the kinematic equations of motion. Let's break down the problem step by step.

Step 1: Determine the vertical and horizontal components of the initial velocity.
The initial velocity of 19.6 m/s at an angle of 30 degrees above the horizontal can be separated into its vertical and horizontal components.

Vertical component: Vsinθ
Horizontal component: Vcosθ

Given:
Initial velocity (V) = 19.6 m/s
Angle (θ) = 30 degrees

Vertical component (Vy) = V * sin(θ)
Vertical component (Vy) = 19.6 * sin(30)
Vertical component (Vy) = 9.8 m/s

Horizontal component (Vx) = V * cos(θ)
Horizontal component (Vx) = 19.6 * cos(30)
Horizontal component (Vx) = 16.96 m/s

Step 2: Determine the time it takes for the ball to reach its highest point.
The time it takes for an object to reach its highest point can be determined using the formula:

Time (t) = Vy / g

Given:
Vertical component (Vy) = 9.8 m/s
Acceleration due to gravity (g) = 9.8 m/s^2

Time (t) = 9.8 / 9.8
Time (t) = 1 second

Step 3: Determine the maximum height of the ball above the ground.
To find the maximum height, we can use the equation of motion:

Vertical displacement (Δy) = Vy * t - 0.5 * g * t^2

Given:
Vertical component (Vy) = 9.8 m/s
Time (t) = 1 second
Acceleration due to gravity (g) = 9.8 m/s^2

Vertical displacement (Δy) = 9.8 * 1 - 0.5 * 9.8 * (1)^2
Vertical displacement (Δy) = 9.8 - 0.5 * 9.8 * 1
Vertical displacement (Δy) = 9.8 - 4.9
Vertical displacement (Δy) = 4.9 meters

Therefore, the maximum height of the ball above the ground is 4.9 meters.

Step 4: Calculate the speed of the ball at the highest point in the trajectory.
The speed at any point in the trajectory can be determined using the formula:

Speed (v) = √ (Vy^2 + Vx^2)

Given:
Vertical component (Vy) = 9.8 m/s
Horizontal component (Vx) = 16.96 m/s

Speed (v) = √ ((9.8)^2 + (16.96)^2)
Speed (v) = √ (96.04 + 287.9616)
Speed (v) = √ 384.0016
Speed (v) = 19.6 m/s

Therefore, the speed of the ball at the highest point in the trajectory is 19.6 m/s.