Physics Projectile Motion Quiz

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Questions and Answers

What is the total time of flight for the cannonball shot at an angle upwards from a 100 m high cliff with a vertical velocity of 20 m/s?

  • 8.5 s
  • 2.5 s
  • 6.5 s
  • 4.5 s (correct)

What is the maximum height reached by the cannonball if it is shot upward at 20 m/s from a height of 100 m?

  • 150 m
  • 120 m
  • 200 m
  • 110 m (correct)

If a projectile is dropped from a height of 100 m, which kinematic equation would correctly calculate the time taken to reach the ground?

  • Ady = ViyAt² + (1/2)ayAt
  • Ady = ViyAt + ayAt²
  • Ady = ViyAt + (1/2)ayAt² (correct)
  • Ady = (1/2)ViyAt² + ayAt

Why is the time variable At always considered positive in projectile motion problems?

<p>Negative time represents an impossible scenario. (B)</p> Signup and view all the answers

In projectile motion problems, what happens to the horizontal component of velocity as the projectile rises and falls?

<p>It remains constant throughout the motion. (B)</p> Signup and view all the answers

What can be concluded about the vertical velocity at the maximum height for a projectile launched upwards?

<p>It is zero momentarily. (A)</p> Signup and view all the answers

When analyzing the range of a projectile shot horizontally from a height, which equation is correctly used?

<p>Adx = VxAt (C)</p> Signup and view all the answers

If the initial vertical velocity of a projectile is negative, what does this indicate about its motion?

<p>It is dropping downwards. (C)</p> Signup and view all the answers

What is the maximum height reached by the cannonball if it is projected with an initial vertical velocity of 20 m/s?

<p>20 m (D)</p> Signup and view all the answers

If the negative sign is interpreted incorrectly in the final velocity of a projectile, what might a student mistakenly conclude about the projectile's motion?

<p>The projectile moves exclusively upwards. (C)</p> Signup and view all the answers

Which kinematic equation is appropriately used to determine the maximum height of a projectile when vertical velocity is involved?

<p>v² = v²1 + 2aΔd (D)</p> Signup and view all the answers

When determining the final velocity of a projectile, what must be considered regarding its x and y components?

<p>The components must be combined using Pythagoras’ theorem. (A)</p> Signup and view all the answers

In the example provided, what is the vertical component of the final velocity when the projectile lands in the net?

<p>22 m/s [down] (B)</p> Signup and view all the answers

What does the term 'horizontal component of velocity' imply for the projectile's motion?

<p>It remains constant due to a lack of horizontal acceleration. (D)</p> Signup and view all the answers

When analyzing the motion of the Great Projecto, what key factor is crucial for calculating the resultant final velocity?

<p>The magnitude and direction of both vector components. (D)</p> Signup and view all the answers

How does the vertical acceleration affect the maximum height a projectile can reach in terms of its vertical motion?

<p>It influences the vertical component of the initial velocity. (A)</p> Signup and view all the answers

What is the maximum height reached by a projectile if its initial vertical velocity is 20 m/s and the acceleration due to gravity is -9.8 m/s²?

<p>40.82 m (C)</p> Signup and view all the answers

What does a negative time value indicate in the context of projectile motion?

<p>The projectile was launched before the reference time. (C)</p> Signup and view all the answers

In the kinematic equation $∆d = v_1∆t + rac{1}{2}ay∆t^2$, what does each term represent?

<p>Change in displacement in either vertical or horizontal direction. (B)</p> Signup and view all the answers

What is the final vertical velocity of a projectile at its maximum height?

<p>0 m/s (B)</p> Signup and view all the answers

In the provided calculations, what was the total horizontal distance traveled by the projectile when considering a time of 7.0 s?

<p>280 m (D)</p> Signup and view all the answers

How does the vertical acceleration affect the projectile’s motion in both the ascent and descent phases?

<p>It causes a decrease in velocity during ascent and an increase during descent. (B)</p> Signup and view all the answers

In analyzing the horizontal and vertical components of projectile motion, what role does the initial horizontal velocity play?

<p>It dictates the maximum horizontal distance traveled. (C)</p> Signup and view all the answers

Which equation is most suitable for finding the time of flight for a projectile with initial vertical velocity of 20 m/s and a vertical displacement of -100 m?

<p>At = (20 m/s ± √(400 + 1960))/ -9.8 m/s² (A)</p> Signup and view all the answers

Flashcards

Projectile Motion

Motion of an object thrown into the air, subject to gravity.

Equations of Motion

Formulas describing x and y motion: Ady=viyat+ayt and Adx=vxAt.

Acceleration in x direction

Always zero because there are no forces acting horizontally.

Acceleration in y direction

Always -9.8 m/s², representing gravity's effect downward.

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Time of Flight

Duration a projectile is in the air, calculated from vertical motion.

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Finding the Range

Horizontal distance traveled by a projectile.

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Example of Range Calculation

A projectile shot horizontally (40 m/s) from 100 m high travels 180 m.

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Projectile at Angle

A projectile shot at an angle travels further than horizontal launches.

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Displacement in y direction

Final position minus initial position in vertical motion.

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Final Velocity Components

Combining x and y velocity vectors to find total final velocity.

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Vertical Final Velocity

Final velocity downward calculated as V2=√(V21+2aΔd).

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Example of Final Velocity

Projectile with initial speeds of 19 m/s x and 23 m/s y lands 2 m up.

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Quadratic Formula

Used to solve equations of the form ax² + bx + c = 0.

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Negative Time

Represents times before the initial time, during upward motion.

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Maximum Height of a Projectile

Point where vertical velocity is zero before falling down.

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Example of Maximum Height

Projectile reaches 20 m above the cliff top before falling.

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Vertical Displacement Equation

Vertical displacement is calculated as Δdy = viyt + 0.5ayt².

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Horizontal Velocity

The constant speed in the x-direction, remains unchanged.

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Projectile Launch Angle

The angle at which a projectile is launched affects its trajectory.

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Gravity's Role in Motion

Gravity pulls down on projectiles, changing their vertical velocity.

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Air Resistance

Force acting opposite to the direction of motion that can affect projectiles.

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Study Notes

Projectile Motion

  • Projectiles follow the same equations in the x and y directions.
  • Projectile motion is described by the following equations: Ady=viyat+aytˇAdy = viyat + ayÅ¥Ady=viyat+aytˇ and Adx=vxAtAdx = vxAtAdx=vxAt.
  • The acceleration in the x direction is always zero.
  • The acceleration in the y direction is always -9.8 m/s².
  • To solve for the range (horizontal distance) of a projectile, you must first solve for the time of flight in the y direction.

Example 3: Finding the Range

  • A projectile shot horizontally (40 m/s) from a 100 m high cliff travels 180 m horizontally.
  • The time of flight is calculated using the equation: At=±√(2∆dy/ay)At = ± √(2∆dy / ay)At=±√(2∆dy/ay).
  • The time of flight is 4.5 s.
  • The range is calculated using the equation: Adx=vxAtAdx = vxAtAdx=vxAt

Example 4: Projection Angled Upward

  • A projectile shot at an angle from a 100 m high cliff travels a greater distance than a projectile shot horizontally.
  • The projectile travels 20 m above the cliff top before falling back down.
  • The displacement in the y direction is defined as the final position minus the initial position.

Finding Final Velocity

  • The final velocity of a projectile can be found by combining the x and y components of the velocity vector.
  • The final velocity in the x direction is the same as the initial velocity because there is no acceleration in the x direction.
  • The final velocity in the y direction can be calculated using the equation: V2=√(V21+2aΔd)V2 = √(V²1 + 2aΔd)V2=√(V21+2aΔd).

Example 6: Finding the Final Velocity

  • A projectile shot with an initial horizontal velocity of 19 m/s and an initial vertical velocity of 23 m/s lands 2.0 m above the point of launch and 70 m away.
  • The final velocity in the y direction is -22 m/s because the projectile is moving downward.
  • The final velocity is found by combining the x and y components of the velocity vector using the head-to-tail method.

Quadratic Formula

  • The quadratic formula is used to solve quadratic equations of the form ax2+bx+c=0ax² + bx + c = 0ax2+bx+c=0.
  • The solution to the quadratic formula is: x=(−b±√(b2−4ac))/(2a)x = (-b ± √(b² - 4ac)) / (2a)x=(−b±√(b2−4ac))/(2a).

Negative Time

  • Negative time in projectile motion can be interpreted as a time earlier in the motion than the initial time.
  • Negative time refers to the earlier part of the parabola where the projectile is moving upwards.

Example 5: Maximum Height of a Projectile

  • The maximum height of a projectile is the point where the vertical velocity is zero.
  • The maximum height can be calculated using the equation: V2=√(V21+2aΔd)V2 = √(V²1 + 2aΔd)V2=√(V21+2aΔd).
  • The maximum height is 20 m above the cliff top in Example 4.

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