A stone is catapulted at time t = 0, with an initial velocity of magnitude 18.6 m/s and at an angle of 41.6° above the horizontal. What are the magnitudes of the (a) horizontal and (b) vertical components of its displacement from the catapult site at t = 1.06 s? Repeat for the (c) horizontal and (d) vertical components at t = 1.76 s, and for the (e) horizontal and (f) vertical components at t = 5.10 s. Assume that the catapult is positioned on a plain horizontal ground.

Note: The Y component of the velocity

decreases with time and is zero at the
maximum height. The velocity increases
as the stone falls.

c. Dx = Xo*t = 13.91 * 1.76 = 24.48 m.

d. Y = Yo + g*Tr = 0
12.35 - 9.8Tr = 0
9.8Tr = 12.35
Tr = 1.26 s. = Rise time or time to reach max ht.

h max = Yo*Tr + 0.5g*t^2 =
12.35*1.26 - 4.9*1.26^2 = 7.78 m.

Tf = 1.76-1.26 = 0.50 s. = Fall time.

h = ho - 0.5g*t^2 = 7.78 - 4.9*0.5^2 =
6.56 m.

e. The stone hits gnd before 5.10 s:
T = 2Tr = 2*1.26 = 2.52 s. = Time in air

Dx = Xo*T = 13.91 * 2.52 = 35.1 m. = Max
range.

f. The stone hits gnd in 2.52 s. So its'
ht. at 5.10 s = 0

To find the magnitudes of the horizontal and vertical components of displacement at different times, we need to use the equations of motion. The horizontal and vertical components can be calculated separately.

Given:
Initial velocity (v0) = 18.6 m/s
Launch angle (θ) = 41.6°
Time (t1) = 1.06 s
Time (t2) = 1.76 s
Time (t3) = 5.10 s

Acceleration due to gravity (g) = 9.8 m/s² (assuming the stone is near the surface of the Earth)

Step 1: Calculate the horizontal component of velocity (v₀x).

The horizontal component of velocity remains constant throughout the motion.
v₀x = v₀ * cos(θ)
v₀x = 18.6 m/s * cos(41.6°)

Step 2: Calculate the horizontal displacement (dx₁) at t = 1.06 s.

The horizontal displacement is given by:
dx = v₀x * t
dx₁ = v₀x * t₁

Step 3: Calculate the vertical component of velocity (v₀y).

The vertical component of velocity changes due to the acceleration due to gravity.
v₀y = v₀ * sin(θ)

Step 4: Calculate the vertical displacement (dy₁) at t = 1.06 s.

The vertical displacement is given by the kinematic equation:
dy = v₀y * t + 0.5 * g * t²
dy₁ = v₀y * t₁ + 0.5 * g * t₁²

Repeat steps 2 to 4 using the respective time values for t = 1.76 s and t = 5.10 s to find the values of dx₂, dy₂, dx₃, dy₃.

Finally, we can calculate the magnitudes of the horizontal and vertical components of displacement:

(a) Magnitude of the horizontal component of displacement at t = 1.06 s:
|dx₁| = abs(dx₁)

(b) Magnitude of the vertical component of displacement at t = 1.06 s:
|dy₁| = abs(dy₁)

(c) Magnitude of the horizontal component of displacement at t = 1.76 s:
|dx₂| = abs(dx₂)

(d) Magnitude of the vertical component of displacement at t = 1.76 s:
|dy₂| = abs(dy₂)

(e) Magnitude of the horizontal component of displacement at t = 5.10 s:
|dx₃| = abs(dx₃)

(f) Magnitude of the vertical component of displacement at t = 5.10 s:
|dy₃| = abs(dy₃)

By following the above steps and plugging in the respective values, you can find the magnitudes of the horizontal and vertical components of displacement at the given times.

To find the magnitudes of the horizontal and vertical components of the displacement, we can use the equations of motion for projectile motion.

The horizontal component of the displacement (x-component) can be found using the formula:

x = v0 * t * cos(θ)

where:
- x is the horizontal displacement
- v0 is the initial velocity magnitude
- t is the time
- θ is the angle of projection

Similarly, the vertical component of the displacement (y-component) can be found using the formula:

y = v0 * t * sin(θ) - (1/2) * g * t^2

where:
- y is the vertical displacement
- v0 is the initial velocity magnitude
- t is the time
- θ is the angle of projection
- g is the acceleration due to gravity (approximately 9.8 m/s^2)

Let's calculate the magnitudes of the horizontal and vertical components of the displacement at each given time:

(a) At t = 1.06 s:
Given:
v0 = 18.6 m/s
θ = 41.6°
t = 1.06 s

Using the equations:
x = v0 * t * cos(θ)
y = v0 * t * sin(θ) - (1/2) * g * t^2

Substituting the given values:
x = 18.6 m/s * 1.06 s * cos(41.6°)
y = 18.6 m/s * 1.06 s * sin(41.6°) - (1/2) * 9.8 m/s^2 * (1.06 s)^2

Now, calculate x and y.

(b) At t = 1.76 s:
Given:
v0 = 18.6 m/s
θ = 41.6°
t = 1.76 s

Using the equations:
x = v0 * t * cos(θ)
y = v0 * t * sin(θ) - (1/2) * g * t^2

Substituting the given values:
x = 18.6 m/s * 1.76 s * cos(41.6°)
y = 18.6 m/s * 1.76 s * sin(41.6°) - (1/2) * 9.8 m/s^2 * (1.76 s)^2

Now, calculate x and y.

(c) At t = 5.10 s:
Given:
v0 = 18.6 m/s
θ = 41.6°
t = 5.10 s

Using the equations:
x = v0 * t * cos(θ)
y = v0 * t * sin(θ) - (1/2) * g * t^2

Substituting the given values:
x = 18.6 m/s * 5.10 s * cos(41.6°)
y = 18.6 m/s * 5.10 s * sin(41.6°) - (1/2) * 9.8 m/s^2 * (5.10 s)^2

Now, calculate x and y.

By solving these equations, you can find the magnitudes of the horizontal and vertical components of the displacement at each given time.

Vo = 18.6m/s[41.6o]

Xo = 18.6*Cos41.6 = 13.91 m/s
Yo = 18.6*sin41.6 = 12.35 m/s

a. X = Xo=13.91 m/s and does not change.
Dx = Xo*t = 13.91 * 1.06 = 14.74 m.

b. Y = Yo + g*t = 12.35 - 9.8*1.06 = 1.962 m/s
h = Yo*t + 0.5g*t^2
Yo=12.35 m/s, t = 1.06 s, g=-9.8 m/s^2.
Solve for h.

c. Same as part "a" except t = 1.76 s.

d. Same as part b except t = 1.76 s.

e. Same as part a except t = 5.10 s.

f. Same as part b except t = 5.10 s.