Verified answer

The vertical part of the velocity is found by the following:

Vy=16.0*sin(30)=8.0 m/s

We must find the time the ball is in the air.

distance to the peak = initial distance + initial velocity* time + 1/2 acceleration * time^2

d= 2+ 8t + 1/2 (-9.81) t^2

At its peak the velocity is zero

Velcoity at the peak = initial velocity + acceleration * time

0 = 8 m/s -9.81 t

t = -8/-9.81 = = 0.82 seconds

d= 2+ 8t + 1/2 (-9.81) t^2 = 2 + 8(0.82) - 1/2 (-9.81)(0.82)^2 = 5.3 meters

time it takes to reach the ground from its peak:

distance = initial velocity * time + 1/2 a t ^2 = 0 + 1/2 *(-9.81)*(t)^2

-5.3 = 0 -1/2 *9.81*t^2 solve for t and you get 1.04 seconds

total time in the air 1.04+0.82 = 1.86 seconds

The x component of this is 16.0 m/s * cos 30 degrees = 13.9 m/s

13.9 m/s * 1.86 seconds = 25.8 m/s

Your teacher should of rounded up.

The horizontal component of the ball’s velocity remains constant.

Horizontal distance = Horizontal component of the velocity * time

Horizontal component of the velocity = 16.0 * cos 30˚ = 13.8564 m/s

Horizontal distance = 16.0 * cos 30˚ * t

We need to determine the total time that the ball is in the air.

As the ball moves upward to its maximum height, its vertical velocity decreases 9.8 m/s each second. At the maximum height, the vertical velocity is 0 m/s. As the ball falls, its vertical velocity increases 9.8 m/s each second.

Time up = (vf – vi) ÷ a

Initial vertical velocity = 16.0 * sin 30˚ = 8 m/s

Acceleration = -9.8 m/s^2

Time up (0 – 8) ÷ -9.8 = 8/9.8

Distance = vi * t + ½ * a * t^2

Distance up = 8 * (8/9.8) + ½ * -9.8 * (8/9.8)^2 ≈ 3.2653 meters

The ball travelled 3.2653 meters upward from its initial position. Since the ball is thrown from a height of 2.00 m above the ground, its maximum height is 5 .2653 meters.

At the maximum height, the vertical velocity is 0 m/s. As the ball falls 5.6253 meters its vertical velocity increases 9.8 m/s each second.

Distance down = 0 * t + ½ * 9.8 * t^2 = 5.2653

4.9 * t^2 = 5.2653

Time down = (5.2653 ÷ 4.9)^0.5

Total time in the air = time up + time down = 8/9.8 + (5.26253 ÷ 4.9)^0.5

Horizontal distance = 16.0 * cos 30˚ * [8/9.8 + (5.2653 ÷ 4.9)^0.5]

Horizontal distance = 25.67 meters

Alternative method

Final velocity^2 = Initial velocity^2 + (2 * acceleration * displacement)

Initial vertical velocity = 8 m/s, Acceleration = -9.8 m/s^3

Displacement = Final height – Initial height

Final height = 0 m, Initial height = 2 m

Displacement = - 2 meters

Final vertical velocity^2 = 8^2 + (2 * -9.8 * -2) = 64 + 39.2 = 103.2

Final vertical velocity = √103.2 m/s

Since the ball is moving downward, it final vertical velocity = -√103.2 m/s

Time = (vf – vi) ÷ a = (-√103.2 – 8) ÷ -9.8 ≈ 1.853 seconds

The ball was in the air for approximately 1.853 seconds.

Horizontal distance = 16.0 * cos 30˚ * [(-√103.2 – 8) ÷ -9.8] = 25.67 meters

I use the alternative method to check the answer.