A ball is thrown from a point 1.1 m above the ground. The initial velocity is 20.0 m/s at an angle of 32.0° above the horizontal.

(a) Find the maximum height of the ball above the ground.

(b) Calculate the speed of the ball at the highest point in the trajectory.

the vertical component of the velocity (Vv)is __ 20.0 * sin(32.0º)

the horizontal component (Vh) is __ 20.0 * cos(32.0º)

the time (T) to max height is __ Vv / g

a) Hmax = (-.5 g T^2) + (Vv * T) + 1.1

b) at Hmax, the ball has no vertical velocity; only horizontal (Vh)

To find the maximum height of the ball above the ground, we can use the equations of motion for projectile motion.

(a) The maximum height occurs when the vertical component of the velocity becomes zero. We can use the equation for vertical displacement to find the maximum height:

Δy = v₀y * t + (1/2) * a * t²

where Δy is the vertical displacement, v₀y is the initial vertical velocity, t is the time, and a is the acceleration due to gravity (-9.8 m/s²).

Initially, the ball is thrown upwards, so the initial vertical velocity, v₀y, can be found using trigonometry:

v₀y = v₀ * sinθ

where v₀ is the initial velocity and θ is the angle above the horizontal.

We can rearrange the equation for vertical displacement to solve for time, t:

Δy = v₀y * t + (1/2) * a * t²
0 = v₀y * t + (1/2) * a * t²

Since the ball reaches maximum height when the vertical velocity becomes zero, we can solve this equation for t:

0 = v₀y * t + (1/2) * a * t²
0 = v₀ * sinθ * t - (1/2) * g * t²

Rearranging and factoring out t:

(1/2) * g * t² = v₀ * sinθ * t

Now we can solve for t:

(1/2) * g * t = v₀ * sinθ
t = (2 * v₀ * sinθ) / g

Plug in the given values:
v₀ = 20.0 m/s
θ = 32.0°
g = 9.8 m/s²

t = (2 * 20.0 * sin(32.0°)) / 9.8

Calculating this, we find t ≈ 3.1 s.

Now, we can plug this value of t into the equation for vertical displacement to find the maximum height:

Δy = v₀y * t + (1/2) * a * t²
Δy = v₀ * sinθ * t + (1/2) * (-9.8) * t²

Δy = 20.0 * sin(32.0°) * 3.1 + (1/2) * (-9.8) * (3.1)²

Calculating this, we find Δy ≈ 11.1 m.

Therefore, the maximum height of the ball above the ground is approximately 11.1 m.

(b) To calculate the speed of the ball at the highest point in the trajectory, we need to find the vertical and horizontal components of the velocity at that point.

At the highest point, the vertical velocity is zero, but the horizontal velocity remains unchanged throughout the motion. Therefore, the speed at the highest point is equal to the horizontal component of the initial velocity, v₀x.

We can find v₀x using the equation:

v₀x = v₀ * cosθ

where v₀ is the initial velocity and θ is the angle above the horizontal.

Using the given values:

v₀ = 20.0 m/s
θ = 32.0°

v₀x = 20.0 * cos(32.0°)

Calculating this, we find v₀x ≈ 17.2 m/s.

Therefore, the speed of the ball at the highest point in the trajectory is approximately 17.2 m/s.