speed 13 m/s ✓ angle = 42° A1. Provide a diagram here A2. the time to peak height of the throw B. The distance above the ground at the peak C. the total flight time D. the horizontal distance travelled by the put 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 * 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
speed 13 m/s ✓ angle = 42° A1. Provide a diagram here A2. the time to peak height of the throw B. The distance above the ground at the peak C. the total flight time D. the horizontal distance travelled by the put 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 * 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
Physics for Scientists and Engineers with Modern Physics
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Author:Raymond A. Serway, John W. Jewett
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Chapter3: Vectors
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Problem 16P: A snow-covered ski slope makes an angle of 35.0 with the horizontal. When a ski jumper plummets onto...
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A shot putter releases a shot at 13 m/s at an angle of 42 degrees to the horizontal and from a height of 1.83 m above the ground. (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:speed 13 m/s
✓ angle = 42°
A1. Provide a diagram here
A2. the time to peak height of the throw
B. The distance above the ground at the peak
C. the total flight time
D. the horizontal distance travelled by the put
![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
* 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%2Fb6092334-c1dd-4337-b46c-a1229f857c13%2Fj39ffc5_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
* 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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