A1. Provide a diagram here A2. the horizontal component of velocity at kick-off B. the vertical component of velocity at kick-off C. the maximum height reached by the ball Only equations to be used: • Vertical (y) direction = Vfy = Viy+ay*t, Vfy = Viy + (g) * t • Position final = Position initial + Velocity initial * time + ½ acceleration * time^2, Horizontal in (x) direction is sfx = S(ix) + V(ix)*t, V(ix) = s(fx) - s(ix)/time or s(fx) = s(ix) = v(ix) * time, vertical (y) direction is s(fy) = s(iy) + v(iy) * time + ½ g * t^2, • • • Vertical direction, displacement is Sfy-Siy = Viy * t + ½ g *t^2, Formula for vertical in (y) direction is Sfy = Siy+Vy* t + ½ g * t^2. Equation for the vertical (y) direction displacement is (Sfy - Siy) = Viy *t+ ½ g*t [Velocity final]^2 = [velocity initial]^2+2 * acceleration * (position final – position initial), formula for this is v2f v2i +2 a (Sf-si) For horizontal x direction formula is Vfx = Vix, for vertical (y) direction formula is V2fy = V2iy + 2g (Sfy Siy), Vertical displacement is (Sfy - Siy)=-vi2iy/2g * Horizontal distance travelled (Range = VH * airborne time) * Projection height = height of release - height of landing * Horizontal distance travelled (Range = VH * airborne time) * t-flight = t peak + t fall * t-peak =-V vertical-initial/g *t fall √-2d/g or square root of -2d/g *d peak height - landing height * Ranget flight * v horizontal *h peak-v2vi/2g
A1. Provide a diagram here A2. the horizontal component of velocity at kick-off B. the vertical component of velocity at kick-off C. the maximum height reached by the ball Only equations to be used: • Vertical (y) direction = Vfy = Viy+ay*t, Vfy = Viy + (g) * t • Position final = Position initial + Velocity initial * time + ½ acceleration * time^2, Horizontal in (x) direction is sfx = S(ix) + V(ix)*t, V(ix) = s(fx) - s(ix)/time or s(fx) = s(ix) = v(ix) * time, vertical (y) direction is s(fy) = s(iy) + v(iy) * time + ½ g * t^2, • • • Vertical direction, displacement is Sfy-Siy = Viy * t + ½ g *t^2, Formula for vertical in (y) direction is Sfy = Siy+Vy* t + ½ g * t^2. Equation for the vertical (y) direction displacement is (Sfy - Siy) = Viy *t+ ½ g*t [Velocity final]^2 = [velocity initial]^2+2 * acceleration * (position final – position initial), formula for this is v2f v2i +2 a (Sf-si) For horizontal x direction formula is Vfx = Vix, for vertical (y) direction formula is V2fy = V2iy + 2g (Sfy Siy), Vertical displacement is (Sfy - Siy)=-vi2iy/2g * Horizontal distance travelled (Range = VH * airborne time) * Projection height = height of release - height of landing * Horizontal distance travelled (Range = VH * airborne time) * t-flight = t peak + t fall * t-peak =-V vertical-initial/g *t fall √-2d/g or square root of -2d/g *d peak height - landing height * Ranget flight * v horizontal *h peak-v2vi/2g
College Physics
1st Edition
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter3: Two-dimensional Kinematics
Section: Chapter Questions
Problem 26PE: A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the...
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A player kicks a football at the start of the game. After a 4 second flight, the ball touches the ground 50 m from the kicking tee. Assume air resistance is negligible and the take-off and landing height are the same (i.e., time to peak = time to fall = ½ total flight time). (Note: For each question draw a diagram to show the vector/s. Show all the step and provide units in the answers. Provide answer to 2 decimal places unless stated otherwise.) Calculate and answer all parts. Only use equations PROVIDED:
Transcribed Image Text:A1. Provide a diagram here
A2. the horizontal component of velocity at kick-off
B. the vertical component of velocity at kick-off
C. the maximum height reached by the ball
![Only equations to be used:
• Vertical (y) direction = Vfy = Viy+ay*t, Vfy = Viy + (g) * t
• Position final = Position initial + Velocity initial * time + ½ acceleration * time^2, Horizontal in (x)
direction is sfx = S(ix) + V(ix)*t, V(ix) = s(fx) - s(ix)/time or s(fx) = s(ix) = v(ix) * time, vertical (y)
direction is s(fy) = s(iy) + v(iy) * time + ½ g * t^2,
•
•
•
Vertical direction, displacement is Sfy-Siy = Viy * t + ½ g *t^2, Formula for vertical in (y) direction is
Sfy = Siy+Vy* t + ½ g * t^2. Equation for the vertical (y) direction displacement is (Sfy - Siy) = Viy
*t+ ½ g*t
[Velocity final]^2 = [velocity initial]^2+2 * acceleration * (position final – position initial), formula for
this is v2f v2i +2 a (Sf-si)
For horizontal x direction formula is Vfx = Vix, for vertical (y) direction formula is V2fy = V2iy + 2g
(Sfy Siy), Vertical displacement is (Sfy - Siy)=-vi2iy/2g
* Horizontal distance travelled (Range = VH * airborne time)
* Projection height = height of release - height of landing
* Horizontal distance travelled (Range = VH * airborne time)
* t-flight = t peak + t fall
*
t-peak
=-V vertical-initial/g
*t fall √-2d/g or square root of -2d/g
*d peak height - landing height
*
Ranget flight * v horizontal
*h peak-v2vi/2g](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14055f04-58ff-4c00-af23-66e4ceb29afb%2Feabf0650-f98c-47eb-ba38-f1cfee00e6b9%2Fhvy6os_processed.png&w=3840&q=75)
Transcribed Image Text:Only equations to be used:
• Vertical (y) direction = Vfy = Viy+ay*t, Vfy = Viy + (g) * t
• Position final = Position initial + Velocity initial * time + ½ acceleration * time^2, Horizontal in (x)
direction is sfx = S(ix) + V(ix)*t, V(ix) = s(fx) - s(ix)/time or s(fx) = s(ix) = v(ix) * time, vertical (y)
direction is s(fy) = s(iy) + v(iy) * time + ½ g * t^2,
•
•
•
Vertical direction, displacement is Sfy-Siy = Viy * t + ½ g *t^2, Formula for vertical in (y) direction is
Sfy = Siy+Vy* t + ½ g * t^2. Equation for the vertical (y) direction displacement is (Sfy - Siy) = Viy
*t+ ½ g*t
[Velocity final]^2 = [velocity initial]^2+2 * acceleration * (position final – position initial), formula for
this is v2f v2i +2 a (Sf-si)
For horizontal x direction formula is Vfx = Vix, for vertical (y) direction formula is V2fy = V2iy + 2g
(Sfy Siy), Vertical displacement is (Sfy - Siy)=-vi2iy/2g
* Horizontal distance travelled (Range = VH * airborne time)
* Projection height = height of release - height of landing
* Horizontal distance travelled (Range = VH * airborne time)
* t-flight = t peak + t fall
*
t-peak
=-V vertical-initial/g
*t fall √-2d/g or square root of -2d/g
*d peak height - landing height
*
Ranget flight * v horizontal
*h peak-v2vi/2g
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