A ball is thrown from a point 1.34 m above the ground. The initial velocity is 18.7 m/s at an angle of 30.8° above the horizontal. Find the maximum height of the ball above the ground.6.02 mYou are correct.Your receipt no. is 162-3367 Previous TriesCalculate the speed of the ball at the highest point in the trajectory.

The speed of the ball at the highest point in the trajectory can be calculated using the initial velocity, the angle of projection, and the acceleration due to gravity.

First, we need to find the vertical component of the initial velocity. This can be calculated using the formula:

Vyi = Vi * sin(θ)

where:

• Vyi is the initial vertical velocity,
• Vi is the initial velocity, and
• θ is the angle of projection.

Substituting the given values:

Vyi = 18.7 m/s * sin(30.8°) = 9.57 m/s

At the highest point in the trajectory, the vertical velocity of the ball is 0 (since it momentarily stops before falling back down). However, the horizontal component of the velocity remains constant throughout the flight. This can be calculated using the formula:

Vxi = Vi * cos(θ)

where:

• Vxi is the initial horizontal velocity.

Substituting the given values:

Vxi = 18.7 m/s * cos(30.8°) = 16.07 m/s

Therefore, the speed of the ball at the highest point in the trajectory is equal to the horizontal component of the velocity, which is 16.07 m/s.