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?

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

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