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: 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) 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
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: 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) 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
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18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter2: Vectors
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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:
![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:
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)
speed 13 m/s
✓ angle = 42°
A1. Provide a diagram here
A2. the time to peak height of the throw](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14055f04-58ff-4c00-af23-66e4ceb29afb%2Fa50c8b31-6b4c-4096-8ad4-1dd9924f0774%2Fazgc8br_processed.png&w=3840&q=75)
Transcribed Image Text: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:
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)
speed 13 m/s
✓ angle = 42°
A1. Provide a diagram here
A2. the time to peak height of the throw
Transcribed Image Text:B. The distance above the ground at the peak
C. the total flight time
D. the horizontal distance travelled by the put
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