A major leaugue baseball player hits a ball 3 feet above the ground with a velocity of 103 feet per secon in the direction of a 10 foot wall that is 300 feet from home plate. If the hit is at an angle of elevation of 34 degrees and there is wind blowing 22 feet per second in the SAME direction horizontally,, determine if the ball clears the fence.

V = 103Ft/s[34o] + 22Ft./s[0o].

Xo = 103*cos34 + 22*cos0 = 107.4 Ft/s.
Yo = 103*sin34 + 22*sin(00 = 57.6 Ft/s.

tan A = Y/X = 57.6/107.4 = 0.53628
A = 28.2o.

V = Xo/cosA = 107.4/cos28.2 = 121.6 Ft/s.[28.2o]

Y^2 = Yo^2 + 2g*h
hmax = ho + (Y^2-Yo^2)/2g
hmax = 3 + (0-(57.6)^2)/-64=54.84 Ft
above gnd.

Tr = (Y-Yo)/g = (0-57.6)/-32 = 1.8 s. =
Rise time.

h = Yo*t + 0.5g*t^2 = 54.84-10
0 + 16t^2 = 44.84
t^2 = 2.8
Tf = 1.67 s. = Time to fall to 10 Ft above gnd.

Range = Xo * (Tr+Tf)=107.4*(1.8+1.67) =
373 Ft. Yes, it clears the fence.

To determine if the ball clears the fence, we need to find the height of the ball when it reaches the wall, considering the angle of elevation, initial velocity, and the effect of wind.

First, we need to break down the initial velocity into its horizontal and vertical components. The horizontal component is the velocity of the ball in the x-direction, unaffected by the wind. The vertical component is the velocity of the ball in the y-direction, affected by gravity.

The horizontal component of the velocity can be found using trigonometry. Since the ball is hit at an angle of elevation of 34 degrees, we can use the sine function to find the horizontal component:

Horizontal velocity (Vx) = Initial velocity (103 ft/s) * sin(angle of elevation)

Vx = 103 ft/s * sin(34 degrees)
Vx = 103 ft/s * 0.559
Vx ≈ 57.377 ft/s

The vertical component of the velocity can also be found using trigonometry. We can use the cosine function to find the vertical component:

Vertical velocity (Vy) = Initial velocity (103 ft/s) * cos(angle of elevation)

Vy = 103 ft/s * cos(34 degrees)
Vy = 103 ft/s * 0.829
Vy ≈ 85.087 ft/s

Now, let's consider the effect of wind. The wind is blowing in the same direction as the ball is hit, so it will add to the horizontal component of the velocity. We need to subtract the wind's velocity from the horizontal component of the velocity:

Effective horizontal velocity (Vx') = Horizontal velocity (Vx) - Wind velocity

Vx' = 57.377 ft/s - 22 ft/s
Vx' = 35.377 ft/s

Now we can analyze the horizontal motion of the ball. We know the horizontal distance from the plate to the wall is 300 feet. We can find the time it takes for the ball to reach the wall:

Time (t) = Distance / Horizontal velocity

t = 300 ft / 35.377 ft/s
t ≈ 8.48 seconds

Next, let's analyze the vertical motion of the ball. The ball is initially 3 feet above the ground, and we want to find the maximum height it reaches before descending. To do this, we can use the motion equation:

Vertical displacement (d) = Initial vertical velocity (Vy) * Time (t) - 0.5 * Acceleration due to gravity (g) * Time (t)^2

Acceleration due to gravity (g) is approximately 32.2 ft/s^2.

d = 85.087 ft/s * 8.48 s - 0.5 * 32.2 ft/s^2 * (8.48 s)^2
d ≈ 364.7 ft

Therefore, the maximum height the ball reaches is approximately 364.7 feet.

Finally, we need to determine if the ball clears the 10-foot wall. To do this, we compare the height of the ball at the wall (which is 3 feet above the ground) and the maximum height it reaches (approximately 364.7 feet).

Since the maximum height is much higher than the height of the wall, we can conclude that the ball clears the fence.

Therefore, the ball clears the 10-foot wall.