A man stands on the roof of a building of height 14.4 m and throws a rock with a velocity of magnitude 28.7 m/s at an angle of 29.9 ∘ above the horizontal. You can ignore air resistance.
a) the maximum height above the roof reached by the rock
b) Calculate the magnitude of the velocity of the rock just before it strikes the ground.
c) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.
the height relative to the ground is
y = 14.4 + (28.7 sin29.9°)t - 4.9t^2
(a) find the vertex of the parabola
(b) find t when y=0; v^2 = ((28.7 sin 29.9°)-9.8t)^2 + ((28.7 cos29.9°)^2)
(c) x = (28.7 cos29.9°)t
To solve this problem, we can break it down into three parts:
a) Finding the maximum height reached by the rock above the roof
b) Calculating the magnitude of the velocity just before it strikes the ground
c) Determining the horizontal distance to the point of impact
Let's start with part (a):
a) To find the maximum height reached by the rock above the roof, we can use the equation for vertical motion. Since we know the initial velocity (28.7 m/s) and the angle of projection (29.9 degrees), we can split the initial velocity into its vertical and horizontal components.
The initial vertical velocity (Vy) can be found by multiplying the initial velocity (V) by the sine of the angle (θ):
Vy = V * sin(θ)
In this case:
V = 28.7 m/s
θ = 29.9 degrees
So we have:
Vy = 28.7 m/s * sin(29.9 degrees)
Next, we use the equation for vertical displacement (s) to calculate the maximum height (h) reached by the rock above the roof:
s = V0y * t + (1/2) * g * t^2
In this case:
s = h = maximum height
V0y = Vy (since there is no vertical acceleration)
t = time taken to reach maximum height
g = acceleration due to gravity (approximately 9.8 m/s^2)
We can rearrange the equation and solve for t when the rock reaches its maximum height:
0 = Vy - g * t
We substitute the value of Vy and g to solve for t:
0 = 28.7 m/s * sin(29.9 degrees) - 9.8 m/s^2 * t
Now, solve for t:
t = [28.7 m/s * sin(29.9 degrees)] / (9.8 m/s^2)
Once we have t, we can find the maximum height (h) using the equation for vertical displacement:
h = V0y * t - (1/2) * g * t^2
Substituting the values:
h = (28.7 m/s * sin(29.9 degrees)) * t - (1/2) * (9.8 m/s^2) * t^2
Now we can calculate h to find the maximum height reached by the rock above the roof.
Moving on to part (b):
b) To calculate the magnitude of the velocity just before the rock strikes the ground, we can use the equation for vertical motion. Since the vertical displacement is equal to the height of the building (14.4 m), we can solve for the final vertical velocity (Vy) when the rock reaches the ground.
Using the equation:
s = V0y * t + (1/2) * g * t^2
Since the rock starts from the maximum height (h) and ends at ground level (s = 0), we can solve for V0y:
0 = Vy * t - (1/2) * g * t^2
Solving for Vy:
Vy = (1/2) * g * t
Now we know Vy just before the rock hits the ground.
Now, let's move on to part (c):
c) To calculate the horizontal distance from the base of the building to the point where the rock strikes the ground, we can use the horizontal component of the initial velocity and the time taken to reach the ground.
The initial horizontal velocity (Vx) can be found by multiplying the initial velocity (V) by the cosine of the angle (θ):
Vx = V * cos(θ)
In this case:
V = 28.7 m/s
θ = 29.9 degrees
So we have:
Vx = 28.7 m/s * cos(29.9 degrees)
Next, we use the equation for horizontal displacement (x) to calculate the distance traveled by the rock:
x = V0x * t
In this case:
x = distance traveled
V0x = Vx (since there is no horizontal acceleration)
t = time taken to reach the ground
Now, using the value of V0x and the time calculated previously, we can find the horizontal distance traveled by the rock.
Remember to use the appropriate units and ensure calculations are performed accurately.