Respuesta :
The motion of the stone thrown by the student, that is acted
upon by gravitational force, is a projectile motion.
Correct responses;
(a) x₀ = 0 m
- y₀ = 49.0 m
(b) v₀ₓ = 15.0 m/s
- [tex]v_{0y}[/tex] = 0 m/s
(c) vₓ = 15.0 m/s
- [tex]v_y[/tex] = 9.8·t
(d) x = 15.0 × t
y = 49.0 - 4.9·t²
(e) Approximately 3.16 seconds
(f) [tex]\vec v_f[/tex] ≈ 34.43
- θ ≈ 64.17° below the horizontal
Which methods are used to analyze the motion of the stone
The given information are;
Magnitude of the velocity with which the student throws the stone, v₀ = 15.0 m/s.
Direction in which the student throws the stone = Horizontally
Height of the cliff from which the student throws the stone = 49.0 m above a horizontal beach
(a) The coordinates of the initial position of the stone are;
- x₀ = The initial location of the stone in relation to the vertical y-axis = 0 m
- y₀ = The initial location of the stone in relation to the horizontal x-axis = 49.0 m
(b) The stone is thrown in the horizontal direction, therefore;
- The component of the initial velocity in the x-direction is v₀ₓ = 15.0 m/s
The component of the initial velocity in the y-direction is zero, therefore;
- [tex]v_{0y}[/tex] = 0 m/s
(c) Given that the force acting on the stone after it is thrown is the (vertical) force of gravity, we have;
The (horizontal) x-component of the velocity is constant, therefore;
vₓ = 15.0 m/s
The vertical y-component of the velocity, [tex]v_y[/tex], is given by formula for accelerated motion as follows;
[tex]v_y[/tex] = [tex]\mathbf{u _y}[/tex] + g·t
Where;
[tex]u_y[/tex] = The initial vertical velocity = 0 m/s
g = Acceleration due to gravity ≈ 9.8 m/s²
t = Time
Therefore;
[tex]v_y[/tex] = 0 + 9.8·t = 9.8·t
[tex]v_y[/tex] = 9.8·t
(d) The position of the stone traveling under constant velocity, in the horizontal direction, x is given as follows;
x = vₓ × t
Therefore;
x = 15.0 × t
The vertical position of the stone, being acted upon by gravity is given by the equation;
y = y₀ - ([tex]u_y[/tex]·t + [tex]\frac{1}{2}[/tex]·g·t²)
Therefore;
y = 49.0 - (0·t - [tex]\frac{1}{2}[/tex] × 9.8 × t² = 49.0 - 4.9·t²
y = 49.0 - 4.9·t²
(e) Taking the beach as the point y = 0, we have;
At the beach, y = 0 = 49.0 - 4.9·t²
Therefore;
At the beach level; 49.0 = 4.9·t²
[tex]t^2 = \dfrac{49}{4.9} = 10[/tex]
t = [tex]\mathbf{\sqrt{10}}[/tex] ≈ 3.16
- The time it takes the stone to strike the beach after below the cliff is approximately 3.16 seconds.
(f) At the level of the beach, we have;
[tex]v_y[/tex] = 9.8 × [tex]\sqrt{10}[/tex] m/s
vₓ = 15.0 m/s
Therefore;
[tex]Velocity \ at \ the \ beach \ level, \ \vec v_f = \mathbf{\sqrt{\left(9.8 \times \sqrt{10} \right)^2 + (15.0)^2}} \approx \underline{34.43}[/tex]
The direction is given as follows;
[tex]tan(\phi) = \mathbf{\dfrac{v_y}{v_x}}[/tex]
Therefore;
[tex]tan(\phi) = \dfrac{9.8 \times \sqrt{10} }{15.0}[/tex]
Given that the vertical motion is in the negative y-direction and
the horizontal motion is in the positive x-direction, the angle of
impact of the stone, ∅, measured clockwise from the negative x-
direction is given as follows;
[tex]The \ angle \ of \ impact, \ \phi= arctan \left(\dfrac{9.8 \times \sqrt{10} }{15.0} \right) \approx 64.17^{\circ}[/tex]
The angle of impact below the horizontal, θ = ∅ ≈ 64.17°
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