Respuesta :
a) The starting speed of the stone was 18.0 m/s, and the cliff's height was 50.0 m.
The expression for the stone's initial x-coordinate
[tex]x_{i}[/tex] = 0
The stone's initial x-coordinate is shown here.
the definition of the top of the cliff's initial y-coordinate
[tex]y_{i}[/tex] = 50m
The stone's initial y-coordination is seen here.
Consequently, the stone's starting position's coordinates are (0, 50.0m)
b) The expression for the stone's starting speed in the x direction
[tex]v_{ix}[/tex] = 18.0m/s
The stone's initial horizontal velocity [tex]v_{ix}[/tex] is seen here. The formula for the starting velocity in the y direction .
[tex]v_{iy}[/tex] = 0m/s
Here, [tex]v_{iy}[/tex] is the stone's initial vertical velocity.
Consequently, the initial component of the stone's speed are [tex]v_{ix}[/tex] = 18.0m/s and [tex]v_{iy}[/tex] = 0m/s.
c) A free fall motion with constant g-force governs the stone's vertical movement.
Particle motion with constant acceleration is what is happening in the y direction.
As a result, the acceleration in the y direction remains constant.
d) Because there is zero acceleration in the x-direction and no net force acting to modify the stone's inertia, the particle's velocity remains constant throughout the motion.
As a result, the motion in the y-direction is motion at a constant speed.
As a result, motion in the x-direction is caused by constant velocity motion.
e) Because the stone doesn't experience any acceleration in the x direction, its speed remains constant throughout the motion.
The stone's ultimate x-direction velocity is equal to its x-direction beginning velocity.
the relationship between the x-final direction's and beginning velocities,
[tex]v_{fx} = v_{ix}[/tex]
Here, [tex]v_{fx}[/tex] is the final horizontal velocity.
The speed in the x direction is independent of time.
The stone experiences a constant acceleration in the y direction, or g, which determines the stone's y-direction velocity.
The formula for the y-ultimate direction's velocity
[tex]v_{fy} = v_{iy} + at[/tex]
Here, [tex]v_{fy}[/tex] is the ultimate vertical velocity, and an is the vertical acceleration.
Substitute 0 for [tex]v_{iy}[/tex] and -g for a in the above equation to find [tex]v_{fy}[/tex]
[tex]v_{fy}[/tex] = -gt
Consequently, the velocity's x and y components are [tex]v_{fx} = v_{ix}[/tex] and [tex]v_{fy}[/tex] = -gt respectively.
f) The term for the stone's x-direction position
[tex]x_{f} = x_{i} + v_{ix}t + a_{x} t^{2}[/tex] (i)
Here, [tex]x_{f}[/tex] is the final horizontal position and [tex]a_{x}[/tex] is the acceleration in the x-direction.
The term for the stone's y-direction position
[tex]y_{f} = y_{i} + v_{iy} t + \frac{1}{2} at^{2}[/tex] (ii)
Here, [tex]y_{f}[/tex] is the final vertical position.
Substitute 0 for [tex]x_{i}[/tex] and 0 for [tex]a_{x}[/tex] in the equation (i)
[tex]x_{f} = 0 + v_{ix} t + 0\\ = v_{ix} t[/tex]
Substitute 0 for [tex]v_{iy}[/tex] and -g for a in the equation (ii)
[tex]y_{f} = y_{i} + 0 - \frac{1}{2} (g)t^{2}[/tex]
[tex]= y_{i} - \frac{1}{2} (g)t^{2}[/tex]
Consequently, the position's x and y components are [tex]v_{xi} t[/tex] and [tex]= y_{i} - \frac{1}{2} (g)t^{2}[/tex] respectively.
g) The formula to calculate time
t = [tex]\sqrt{\frac{2h}{g} }[/tex]
Substitute 50m for h and 9.8m/[tex]s^{2}[/tex] for g in above equation to find t.
t = [tex]\sqrt{\frac{2(50m)}{9.8m/s^{2} } }[/tex]
= 3.19s
As a result, the stone will hit the sea below the cliff in 3.19 seconds.
h) The expression to calculate velocity.
[tex]v_{fy}[/tex] = -gt
Substitute 3.19s for t and 9.8 m/[tex]s^{2}[/tex] Obtain for g in the previous equation [tex]v_{fy}[/tex].
[tex]v_{fy}[/tex] = (9.8m/[tex]s^{2}[/tex]) (3.19s)
= -31.26m/s
≈ - 31.3m/s
The method for estimating the speed of stone land
v = [tex]\sqrt{v^{2} _{fx} }+ \sqrt{v^{2} _{fx} }[/tex] (iii)
The speed of the stone when it lands in this case is v.
The formula for calculating the stone land's angle
∅ = [tex]tan^{-1}[/tex] ( [tex]\frac{v_{fy} }{v_{fx} } )[/tex] (iv)
Substitute 18m/s for [tex]v_{fx}[/tex] and -31.3m/s for [tex]v_{fy}[/tex] in equation (iii) to find v.
v = [tex]\sqrt{(18m/s)^{2} }[/tex] + [tex]\sqrt{(-31.3m/s)^{2} }[/tex]
= 36.1m/s
As a result, the stone was moving at 36.1 m/s when it hit the ground.
Substitute 18m/s for [tex]v_{fx}[/tex] and -31.3m/s for [tex]v_{fy}[/tex] in equation (iv) to find ∅.
∅ = [tex]tan^{-1}[/tex] ([tex]\frac{-31.3m/s}{18m/s} )[/tex]
= -60.09°
≈ -60.1°
As a result, the stone terrain has an angle that is -60.1° below the horizontal.
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