A ball is thrown from a cliff upward with a speed of 20 m/s, the angle is 33 degrees and the ball hits the ground 9.2 seconds later, how high is the cliff to the nearest meter?
Is the 33 degrees with respect to vertical or horizontal? You need the initial vertical velocity component to answer the question. Call it Vyo.
Vyo*t - (g/2) t^2 = -H
Solve for the cliff height H. You know what t is: 9.2 s.
To find the height of the cliff, we need to break down the motion of the ball into its horizontal and vertical components.
First, let's analyze the vertical component of the motion. The ball is initially thrown upwards with a speed of 20 m/s making an angle of 33 degrees with the horizontal axis. Let's consider the upward direction as positive.
We can use the following kinematic equation to calculate the vertical displacement (change in height) of the ball:
y = v₀ * t + (1/2) * a * t^2
where:
y = vertical displacement (height of the cliff)
v₀ = initial vertical velocity (20 m/s * sin(33°))
t = time (9.2 seconds)
a = acceleration due to gravity (-9.8 m/s^2, taking downward direction as negative)
Plug in the given values into the equation to find the vertical displacement:
y = (20 * sin(33°)) * 9.2 + (1/2) * (-9.8) * (9.2)^2
Calculating this equation will give us the vertical displacement or the height of the cliff.
Now let's solve it step by step:
1. Calculate the initial vertical velocity (v₀):
v₀ = 20 m/s * sin(33°)
2. Plug in the known values in the equation:
y = (v₀ * t) + (1/2) * a * t^2
3. Substitute values into the equation and solve for y:
y = [(20 * sin(33°)) * 9.2] + (1/2) * (-9.8) * (9.2)^2
Evaluating this equation will give us the vertical displacement, which is the height of the cliff. Rounding the answer to the nearest meter will give us the final result.