A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed
1 answer:
Answer:
the rate of the change of the length of the shadow is - 0.8625 m/s.
The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.
Explanation:
Given the data in the question;
Let x represent the man's distance from building,
initially x = 1m2
dx/d t= -2.3 m/s
Also Let y represent shadow height
so we determine dy/dt when x is 4m from the building
form the image description of the problem, we see two-like triangles with the same base and height ratios
so
2 / (12-x) = y / 12
24 = y(12 - x )
y = 24 / (12-x)
dy/dt = 24/(12-x)² × dx/dt
Now at x = 4,
we substitute
dy/dt will be;
⇒ 24/(12 - 4)² × -2.3
= 24/64 - 2.3
= 0.375 × -2.3
dy/dt = - 0.8625 m/s
Therefore, the rate of the change of the length of the shadow is - 0.8625 m/s.
The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.
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The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.
a) the position are
time (s) x (m) y(m)
0 0 0
2.0 3.6 1.2
3.0 5.4 0.9
b) The aceleration is g = 0.6 m / s²
c) The launch angle θ = 33.7º
given parameters
- the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s
- the movement times t = 1.0s, 2.0s and 3.0 s
to find
a) position
b) acceleration
c) launch angle
Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration
b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.
v_y = v_{oy} - g t
where v and v({oy} are the velocities of the body, g the acceleration of the planet's gravity and t the time
0 = v_{oy} - gt
g = v_{oy} / t
from the graph we observe that the highest point occurs for t = 2.0 s
g = 1.2 / 2.0
g = 0.6 m / s²
a) The position is requested for several times
X axis
in this axis there is no acceleration so we can use the uniform motion relationships
vₓ = x / t
x = vₓ t
where x is the position, vx is the velocity and t is the time
we calculate for the time
t = 0.0 s
x₀ = 0
t = 2.0 s
x₂ = 1.8 2
x₂ = 3.6 m
t = 3.0 s
x₃ = 1.8 3
x₃ = 5.4 m
Y axis
In this axis there is the acceleration of the planet, let us use for the position the relation
y = v_{oy} t - ½ g t²
t = 0.0 s
y₀ = 0
y₀ = 0 m
t = 2.0 s
y₂ = 1.2 2 - ½ 0.6 2²
y₂ = 1.2 m
t = 3.0 s
y₃ = 1.2 3 - ½ 0.6 3²
y₃ = 0.9 m
c) the launch angle use the trigonometry relation
tan θ =
θ = tan⁻¹
θ = tan⁻¹
θ = 33.7º
measured counterclockwise from the positive side of the x-axis
With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.
a) the position are
time (s) x (m) y(m)
0 0 0
2.0 3.6 1.2
3.0 5.4 0.9
b) The aceleration is g = 0.6 m / s²
c) The launch angle θ = 33.7ºto)
learn more about projectile launch here:
Giants.
- thanks
- hope it helps.