A man stands on the roof of a 15.4 m tall building and throws a rock with a speed of 29.0 m/s at an angle of 33.3 degrees above the horizontal. Calculate
A) the maximum height above the roof that the rock reaches
B) the speed of the rock just before it strikes the ground; and
C) the horizontal range from the base of the building to the point where the rock strikes the ground.
free fall equation ... h = -1/2 g t^2 + [29.0 sin(33.3º) t] + 15.4 m
the max height occurs at the vertex of the parabola ... on the axis of symmetry
... correct for the height of the building
the horizontal speed is constant ... 29.0 cos(33.3º) m/s
the vertical speed at impact is ... √(2 g 15.4) m/s downward
the two speed components add as vectors ... a^2 + b^2 = c^2
the range is the horizontal speed divided by the time of flight
... plug in zero for h in the free fall equation to find t
Vo = 29m/s[33.3o].
Xo = 29*Cos33.3 = 24.2 m/s.
Yo = 29*sin33.3 = 15.9 m/s.
A. Y^2 = Yo^2 + 2g*h = 0,
15.9^2 + (-19.6)h = 0,
h = 12.9 m.
B. Y^2 = Yo^2 + 2g*h = 0 + 19.6 * (12.9+15.4) = 554.7,
Y = 23.6 m/s.
V = sqrt(Xo^2 + Y^2) = sqrt(24.2^2 + 23.6^2) =
C. Y = Yo + g*Tr = 0,
15.9 + (-9.8)Tr = 0,
Tr = 1.62 s. = Rise time.
0.5g*Tf^2 = (15.4 + 12.9),
Tf = 2.40 s. = Fall time.
Range = Xo(Tr + Tf)
Range = sqrt(Xo^2 + Y^2) = sqrt(24.2^2 + 23.6^2) =
To solve this problem, we can use the following equations of motion:
1) Vertical motion:
- Final vertical velocity (v_y) = Initial vertical velocity (v_y0) + (acceleration due to gravity * time)
- Final height (h) = Initial height (h_0) + (Initial vertical velocity * time) + (0.5 * acceleration due to gravity * time^2)
2) Horizontal motion:
- Final horizontal velocity (v_x) = Initial horizontal velocity (v_x0)
- Horizontal displacement (d) = Initial horizontal velocity * time
Given:
Initial height (h_0) = 15.4 m
Initial vertical velocity (v_y0) = 29.0 m/s * sin(33.3°)
Acceleration due to gravity (g) = 9.8 m/s^2
Angle (θ) = 33.3°
A) To find the maximum height above the roof that the rock reaches:
1) Calculate the time it takes for the rock to reach its maximum height:
Initial vertical velocity (v_y0) = 29.0 m/s * sin(33.3°)
v_y = 0 m/s at maximum height
Use the equation: v_y = v_y0 + (g * time)
Substitute the values, solve for time.
2) Calculate the maximum height (h):
Use the equation: h = h_0 + (v_y0 * time) + (0.5 * g * time^2)
Substitute the values, solve for h.
B) To find the speed of the rock just before it strikes the ground:
1) Calculate the time it takes for the rock to hit the ground:
Use the equation: h = h_0 + (v_y0 * time) + (0.5 * g * time^2)
Substitute h = 0, solve for time.
2) Calculate the final vertical velocity (v_y) at ground level:
Use the equation: v_y = v_y0 + (g * time)
Substitute the values, solve for v_y.
3) Calculate the final speed (v) at ground level:
Use the equation: v = sqrt(v_x^2 + v_y^2)
Substitute v_x = initial horizontal velocity, and v_y, solve for v.
C) To find the horizontal range (d):
1) Calculate the time it takes for the rock to hit the ground:
Use the equation: h = h_0 + (v_y0 * time) + (0.5 * g * time^2)
Substitute h = 0, solve for time.
2) Calculate the final horizontal velocity (v_x) at ground level:
Use the equation: v_x = v_x0
Substitute the value, solve for v_x.
3) Calculate the horizontal range (d) from the base of the building:
Use the equation: d = v_x * time
Substitute the values, solve for d.
Now let's calculate the values step by step.
To solve the given problem, we can use the equations of motion to find the maximum height, speed just before striking the ground, and horizontal range of the rock.
Before we start, let's break down the initial velocity of the rock into its vertical and horizontal components.
Vertical component: V₀y = V₀ * sin(θ)
Horizontal component: V₀x = V₀ * cos(θ)
Where:
V₀ is the initial speed of the rock (29.0 m/s).
θ is the angle of projection above the horizontal (33.3 degrees).
A) To find the maximum height above the roof that the rock reaches (hmax), we can use the equation of motion:
Vf² = Vi² + 2 * a * d
At the maximum height, the final vertical velocity (Vf) is 0 m/s, the initial vertical velocity (Vi) is V₀y, the acceleration (a) is -9.8 m/s² (due to gravity), and the displacement (d) is the maximum height (hmax).
0 = V₀y² + 2 * (-9.8) * hmax
Solving for hmax:
0 = (V₀ * sin(θ))² + 2 * (-9.8) * hmax
B) To find the speed of the rock just before it strikes the ground (vf), we can use the equation of motion:
vf² = vi² + 2 * a * d
At the ground, the final vertical velocity (vf) is unknown, the initial vertical velocity (vi) is V₀y, the acceleration (a) is -9.8 m/s², and the displacement (d) is the height of the building (15.4 m).
vf² = (V₀ * sin(θ))² + 2 * (-9.8) * 15.4
C) To find the horizontal range (R) from the base of the building to the point where the rock strikes the ground, we can use the equation of motion:
R = V₀x * t
Where t is the time of flight, which is given by:
t = 2 * (V₀ * sin(θ)) / g
Where g is the acceleration due to gravity (9.8 m/s²).
Solving these equations will give us the answers to parts A, B, and C of the problem.