A stone is aimed at a cliff of height h with an initial speed of v = 36.0 m/s directed 75.0° above the horizontal, as shown in the Figure below by the arrow. The stone strikes at A, 6.25 s after launching. What is the height of the cliff?
To find the height of the cliff (h), we can use the equations of projectile motion.
First, let's analyze the motion in the horizontal (x) direction. The horizontal velocity (Vx) remains constant throughout the motion. We can find Vx using the formula:
Vx = v * cosθ
where:
v = initial speed = 36.0 m/s
θ = launch angle = 75.0°
Vx = 36.0 * cos(75.0°)
Vx ≈ 9.19 m/s
Next, let's analyze the motion in the vertical (y) direction. The vertical velocity (Vy) changes due to gravity. We can find Vy using the formula:
Vy = v * sinθ
Vy = 36.0 * sin(75.0°)
Vy ≈ 34.78 m/s
Now, we can find the time it takes for the stone to reach point A (6.25 seconds). Since the time of flight is the same for both the x and y directions, we can use the y-direction to find the time of flight (t) using the formula:
y = Vy * t + (1/2) * g * t^2
where:
y = direction in the y-direction = height of the cliff = h
g = acceleration due to gravity ≈ 9.8 m/s^2
h = Vy * t - (1/2) * g * t^2
Rearranging the equation, we get:
(1/2) * g * t^2 - Vy * t + h = 0
We know the time of flight (t = 6.25 s), so substituting the values:
(1/2) * 9.8 * (6.25)^2 - 34.78 * 6.25 + h = 0
Finally, solving the equation for h will give us the height of the cliff.