Physics, 20.12.2019 13:31 nardosem9

Ashot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. the shot hits the ground 2.08 s later. you can ignore air resistance.

part a

what is the x-component of the shot's acceleration while in flight?

part b

what is the y-component of the shot's acceleration while in flight?

part c

what is the x-component of the shot's velocity at the beginning of its trajectory?

part d

what is the y-component of the shot's velocity at the beginning of its trajectory?

part e

what is the x-component of the shot's velocity at the end of its trajectory?

part f what is the y-component of the shot's velocity at the end of its trajectory?

Answers: 1

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Answer from: econsta3

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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Answer from: Raync8780

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:

F=mg

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:

a_y = g = -9.8 m/s^2

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

$v_x = v_0 cos \theta$

where

v_0 = 12.0 m/s is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s$

D) 9.3 m/s

The y-component of the shot's velocity is given by:

$v_y = v_0 sin \theta$

where

v_0 = 12.0 m/s is the initial velocity

$\theta=51.0^{\circ}$ is the angle of the shot

Substituting into the equation, we find

$v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s$

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:

v_x = 7.6 m/s

F) -11.1 m/s

The y-component of the velocity at time t is given by:

v_y(t) = v_y + at

where

v_y = 9.3 m/s 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:

v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s

where the negative sign means the vertical velocity is now downward.

Answer from: Quest

Molybdeum(mo)orruthenium (ru)actually, elements with similar chemical properties as technetium would bemn, re, and bh. this is because elements in the same group/family (column) have similar chemical properties. elements in the same period (row) have the same number of energy levels (or shells).

Answer from: Quest

Answertraveling around a circular trackexplanation"traveling around a circular track" is the best answer as you could be traveling around a circular track at a constant speed. your velocity would therefore have the same magnitude but will always be constantly direction. since acceleration is the rate of change of velocity, the rate of change will only involve the direction of velocity and not the magnitude.wrong choices: "an object at rest" does not accelerate at all"increasing speed while traveling around a curve" means that the acceleration will not be constant due to only change in direction; it will be constant due to both magnitude and direction"an object in free fall" does not necessarily change direction

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Ashot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51....

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