Respuesta :
A) Zero
The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.
B) -9.8 m/s^2
The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:
[tex]F=mg[/tex]
where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:
[tex]a_y = g = -9.8 m/s^2[/tex]
where the negative sign means it points downward.
C) 7.6 m/s
The x-component of the shot's velocity is given by:
[tex]v_x = v_0 cos \theta[/tex]
where
[tex]v_0 = 12.0 m/s[/tex] is the initial velocity
[tex]\theta=51.0^{\circ}[/tex] is the angle of the shot
Substituting into the equation, we find
[tex]v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s[/tex]
D) 9.3 m/s
The y-component of the shot's velocity is given by:
[tex]v_y = v_0 sin \theta[/tex]
where
[tex]v_0 = 12.0 m/s[/tex] is the initial velocity
[tex]\theta=51.0^{\circ}[/tex] is the angle of the shot
Substituting into the equation, we find
[tex]v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s[/tex]
E) 7.6 m/s
We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:
[tex]v_x = 7.6 m/s[/tex]
F) -11.1 m/s
The y-component of the velocity at time t is given by:
[tex]v_y(t) = v_y + at[/tex]
where
[tex]v_y = 9.3 m/s[/tex] is the initial y-velocity
a = g = -9.8 m/s^2 is the vertical acceleration
t is the time
Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:
[tex]v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s[/tex]
where the negative sign means the vertical velocity is now downward.
A) The x-component of the shot's acceleration while in flight = 0 m/s²
B) The y-component of the shot's acceleration while in flight = 9.33 m/s²
C) The x-component of the shot's acceleration at the beginning of its trajectory; uₓ = 7.55 m/s
D) The y-component of the shot's acceleration at the beginning of its trajectory; u_y = 9.33 m/s
E) The x-component of the shot's velocity at the end of its trajectory is;
uₓ = = 7.55 m/s
F) The y-component of the shot's velocity at the end of its trajectory is; v = -11.05 m/s
We are given;
Initial velocity of shot; u = 12 m/s
Angle of shot; θ = 51°
time taken to hit the ground; t = 2.08 s
A) Since this is a projectile motion, it means that the acceleration in place is acceleration due to gravity. Now, gravity acts in only the vertical direction which is the y-component. Thus, there will be no x-component acceleration.
∴ x-component acceleration = 0 m/s²
B) We established in A above that there is a vertical component of acceleration due to gravity.
Now, the shot object is going upwards and as a result the acceleration due to gravity will act in the opposite direction which is downwards.
Thus; y-component of acceleration = -9.8 m/s²
C) Formula for the x-component of velocity at the beginning is;
uₓ = u cos θ
∴ uₓ = 12 cos 51°
uₓ = 7.55 m/s
D) Formula for the y-component of velocity at the beginning is;
u_y = u sin θ
∴ u_y = 12 sin 51°
u_y = 9.33 m/s
E) Since there is no acceleration acting in the x-direction, it means that x-component of velocity at the beginning will remain constant at the end.
Thus;
x-component of velocity at the end is; uₓ = 7.55 m/s
F) The y-component velocity at the end is given by the formula;
v = u + gt
plugging in the relevant values;
v = 9.33 + (-9.8*2.08)
v = -11.05 m/s
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