A stone is thrown with an initial speed of 15 m/s at an angle of 37 above the horizontal from the top of a 35 m building. If g = 9.8 m/s2 and air resistance is negligible, then what is the magnitude of the vertical velocity component of the rock as it hits the ground?

First calulate the time T when it hits the ground.

y = 35 + Vo sin37*T - (g/2)T^2 = 0

Solve that for time T. You know what Vo and g are. You must solve a quadratic. Take the positive root.

Get the vertical velocity component by solving
Vy (at t = T) = Vo sin 37 - g*T

It should end up being a negative number.

To find the magnitude of the vertical velocity component of the rock as it hits the ground, we need to find the vertical velocity just before it hits the ground.

We can break down the initial velocity into horizontal and vertical components.

Given:
Initial speed (v₀) = 15 m/s
Angle above the horizontal (θ) = 37°
Height of the building (h) = 35 m
Acceleration due to gravity (g) = 9.8 m/s²

First, let's find the initial vertical velocity component (v₀y):
v₀y = v₀ * sin(θ)
v₀y = 15 * sin(37°)
v₀y = 15 * 0.6018
v₀y ≈ 9.03 m/s

Next, let's find the time it takes for the rock to fall from the top of the building to the ground. We can use the equation:
h = (1/2) * g * t²

Rearranging the equation to solve for t:
t = sqrt((2h)/g)
t = sqrt((2 * 35) / 9.8)
t = sqrt(7.14)
t ≈ 2.67 s

Since the downward vertical acceleration due to gravity is constant, the time taken to fall from the top of the building to the ground is the same time it takes for the rock to reach the ground.

Finally, we can find the final vertical velocity component (vfy) using the equation:
vfy = v₀y + g * t
vfy = 9.03 + 9.8 * 2.67
vfy ≈ 34.31 m/s

Therefore, the magnitude of the vertical velocity component of the rock as it hits the ground is approximately 34.31 m/s.

To find the magnitude of the vertical velocity component of the stone as it hits the ground, we need to analyze the motion of the stone.

We can start by determining the time it takes for the stone to hit the ground. Since we know the height of the building, we can use the equation for vertical displacement:

Δy = v₀t + (1/2)gt²

Where:
Δy = vertical displacement (35 m)
v₀ = initial vertical velocity component
t = time
g = acceleration due to gravity (-9.8 m/s²)

Since the stone is thrown at an angle, it has both horizontal and vertical velocity components. The initial vertical velocity component (v₀) can be calculated by multiplying the initial speed (15 m/s) by the sine of the launch angle (37°):

v₀ = 15 m/s * sin(37°)

Now we can substitute the values into the equation:

35 m = (15 m/s * sin(37°))t + (1/2)(-9.8 m/s²)t²

We can rearrange the equation to form a quadratic equation in terms of t:

(1/2)(-9.8 m/s²)t² + (15 m/s * sin(37°))t - 35 m = 0

Since we are interested in the time it takes for the stone to hit the ground, we only consider the positive value of t (ignoring the negative value).

Using the quadratic formula, we can solve for t:

t = (-b + √(b² - 4ac)) / (2a)

Where:
a = (1/2)(-9.8 m/s²) = -4.9 m/s²
b = (15 m/s * sin(37°))
c = -35 m

Once we have the value of t, we can use it to find the vertical velocity component of the stone just before it hits the ground. We can use the equation:

v = v₀ + gt

Since the stone is falling downwards, the final vertical velocity component (v) would be negative.

Substituting the known values:

v = (15 m/s * sin(37°)) + (-9.8 m/s²)(t)

Finally, you can input the value of t into the equation to find the magnitude of the vertical velocity component of the stone as it hits the ground.