A ball is thrown from a point 1.38 m above the ground. The initial velocity is 19.9 m/s at an angle of 33.8° 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.

a. hmax = ho + (Y^2-(Vo*sin33.8)^2)/2g.

hmax = 1.38 + (0-122.6)/19.6 = 6.26 m.

b. Velocity is zero at max. ht.

To find the maximum height of the ball above the ground, we can first calculate the time it takes for the ball to reach its highest point, and then use that time to find the maximum height.

1. Calculate the vertical component of the initial velocity: vy = v * sin(θ)
where v is the initial velocity and θ is the angle above the horizontal.
vy = 19.9 m/s * sin(33.8°) = 10.8 m/s

2. Since the ball is thrown upwards, the vertical component of the acceleration is -9.8 m/s^2 (due to gravity).

3. Use the kinematic equation: vy = vy0 + a * t
where vy is the final vertical velocity, vy0 is the initial vertical velocity (vy), a is the vertical acceleration (-9.8 m/s^2), and t is the time.
0 = 10.8 m/s - 9.8 m/s^2 * t (at the highest point, vy = 0)
Solving for t, we get t = 10.8 m/s / 9.8 m/s^2 ≈ 1.10 s.

4. Now, we can find the maximum height using the equation: y = y0 + vy0 * t + 0.5 * a * t^2
where y is the final height, y0 is the initial height (1.38 m), vy0 is the initial vertical velocity (10.8 m/s), a is the vertical acceleration (-9.8 m/s^2), and t is the time.
y = 1.38 m + 10.8 m/s * 1.10 s + 0.5 * (-9.8 m/s^2) * (1.10 s)^2 ≈ 6.14 m

Therefore, the maximum height of the ball above the ground is approximately 6.14 meters.

To calculate the speed of the ball at the highest point in the trajectory, we can use the equation:
v = sqrt(vx^2 + vy^2)
where vx is the horizontal component of the velocity and vy is the vertical component of the velocity at the highest point.

1. Calculate the horizontal component of the velocity: vx = v * cos(θ)
where v is the initial velocity and θ is the angle above the horizontal.
vx = 19.9 m/s * cos(33.8°) = 16.6 m/s

2. At the highest point, vy = 0 m/s.

3. Use the equation: v = sqrt(vx^2 + vy^2)
v = sqrt((16.6 m/s)^2 + (0 m/s)^2) = 16.6 m/s

Therefore, the speed of the ball at the highest point in the trajectory is 16.6 m/s.