A baseball is thrown off a cliff at an angle of 57 degrees with a speed of 25 m/s. How high is the cliff to the nearest tenth of a meter, if the ball hits the ground 6.2 sec later.

y = h + 25cos57 t - 4.9t^2

So, if takes 6.2 sec to hit at y=0, then

h + 13.6*6.2 - 4.9*6.2^2 = 0
h = 104.0m

the answer to the question is 58 why is it not matching up?

Ahh. I used the horizontal component of v, not the vertical. Chane cos to sin and things should work out.

Shoulda caught that, guy...

Thank you

To find the height of the cliff, we need to analyze the motion of the baseball in two dimensions: horizontal and vertical.

First, let's break down the initial velocity of the baseball into its horizontal and vertical components using trigonometry. The horizontal component (Vx) remains constant, while the vertical component (Vy) changes due to the effect of gravity.

Given that the angle of projection is 57 degrees and the initial speed is 25 m/s:

Vx = V * cos(θ) = 25 m/s * cos(57°)
Vy = V * sin(θ) = 25 m/s * sin(57°)

Calculating these values:

Vx = 25 m/s * cos(57°) ≈ 13.3 m/s
Vy = 25 m/s * sin(57°) ≈ 21.8 m/s

Now, let's consider the motion of the baseball in the vertical direction. Using the formula for the displacement of an object in free fall:

Δy = (Vyi * t) + (1/2) * g * t^2

Where:
Δy = vertical displacement (height of the cliff)
Vyi = initial vertical velocity component (Vy)
t = time of flight (6.2 sec)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Plugging in the values:

Δy = (21.8 m/s * 6.2 sec) + (1/2) * 9.8 m/s^2 * (6.2 sec)^2

Calculating this:

Δy = 135.16 m + 1/2 * 9.8 m/s^2 * 38.44 sec^2

Δy ≈ 135.16 m + 18.9 m ≈ 154.06 m

Therefore, the height of the cliff is approximately 154.1 meters to the nearest tenth of a meter.