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Physics for Electrical and Electronics Engineering | Projectile Motion, Assignments of Physics Fundamentals

Practice Problem sets with solutions in the course subject Physics for Electrical and Electronics Engineering

Typology: Assignments

2018/2019

Available from 02/20/2022

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Problem Set 1: Projectile Motion
Direction: Answer the following problems correctly. Show your complete solutions.
1. A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. How long does it take
the stone to reach the bottom of the cliff? How far from the base of the cliff does the stone strike the ground?
Given: Vx = 5.0 m/s, y = 78.4m, g = โˆ’9.8๐‘š/๐‘ 2
Unknown: t, x
Solution: ๐‘ก = โˆšโˆ’2๐‘ฆ
๐‘” x = ๐‘‰๐‘ฅ๐‘ก
๐‘ก = โˆšโˆ’2(78.4๐‘š)
โˆ’9.8๐‘š/๐‘ 2 = 4s x = 5.0๐‘š/๐‘  (4๐‘ ) = 20ms
Answer: t = 4s, x = 20m
2. A player kicks a football from the ground level with a velocity 27.0 m/s at an angle of 300 above the horizontal.
Find its โ€˜hang timeโ€™, horizontal distance and its maximum height.
Given: Vo =27 m/s, ฮธ=30โˆ˜, g =9.8๐‘š/๐‘ 2
Unknown: T, R=x, ๐‘Œ๐‘š๐‘Ž๐‘ฅ
Solution: ๐‘‡ = 2๐‘ฃ๐‘œ๐‘ ๐‘–๐‘›๐œƒ
๐‘” ๐‘… = ๐‘ฃ๐‘œ๐‘ ๐‘–๐‘›2๐œƒ
2๐‘” ๐‘Œ = ๐‘ฃ๐‘œ
2๐‘ ๐‘–๐‘›2๐œƒ
2๐‘”
๐‘‡ = 2(27๐‘š
๐‘ )๐‘ ๐‘–๐‘›30ยฐ
9.8๐‘š/๐‘ 2 = 2.75s ๐‘… = (27๐‘š/๐‘ )2sin(2โˆ—30ยฐ)
9.8๐‘š/๐‘ 2 = 64.422m ๐‘Œ = (27๐‘š/๐‘ )2๐‘ ๐‘–๐‘›230ยฐ
2(9.8๐‘š
๐‘ 2) = 9.298m
Answer: T = 2.75s, R = 64.42m, T = 9.3m
3. A long jumper leaves the ground with an initial velocity of 10.5 m/s at an angle of 25-degrees above the
horizontal. What is the time of flight, the horizontal distance, and the peak height of the long-jumper?
Given: Vo = 10.5 m/s, =25ยฐ , 2๐œƒ = 50ยฐ, g = 9.8๐‘š/๐‘ 2
Unknown: ๐‘ก๐‘“๐‘™๐‘–๐‘”โ„Ž๐‘ก, R=x, ๐‘Œ๐‘š๐‘Ž๐‘ฅ
Solution: ๐‘… = ๐‘ฃ๐‘œ๐‘ ๐‘–๐‘›2๐œƒ
2๐‘” ๐‘ก๐‘“๐‘™๐‘–๐‘”โ„Ž๐‘ก = ๐‘ฅ
๐‘‰๐‘œ๐‘๐‘œ๐‘ ๐œƒ ๐‘Œ๐‘š๐‘Ž๐‘ฅ = ๐‘‰๐‘œ๐‘ ๐‘–๐‘›๐œƒ๐‘กโˆ’ 1
2๐‘”๐‘ก2
๐‘… = (10.5๐‘š/๐‘ )2sin 50ยฐ
9.8๐‘š/๐‘ 2 = 8.62m = x ๐‘ก๐‘“๐‘™๐‘–๐‘”โ„Ž๐‘ก = 8.62๐‘š
10.5๐‘š
๐‘ cos (25ยฐ) = 0.91s
๐‘Œ๐‘š๐‘Ž๐‘ฅ =10.5๐‘š
๐‘ ๐‘ ๐‘–๐‘›(25ยฐ)(0.46) โˆ’1
2(9.8๐‘š/๐‘ 2)(0.46)2 = 1.00m
Answer: R = x = 8.62m, ๐‘ก๐‘“๐‘™๐‘–๐‘”โ„Ž๐‘ก =0.91๐‘  , ๐‘Œ๐‘š๐‘Ž๐‘ฅ = 1.00๐‘š
4. During World War I, the Germans constructed an enormous railway gun that was used to bombard Paris from
a distance of 120 km.
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Problem Set 1: Projectile Motion Direction: Answer the following problems correctly. Show your complete solutions.

  1. A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m high. How long does it take the stone to reach the bottom of the cliff? How far from the base of the cliff does the stone strike the ground? Given: Vx = 5.0 m/s, y = 78.4m, g = โˆ’ 9. 8 ๐‘š/๐‘ ^2 Unknown: t, x Solution: ๐‘ก = โˆš โˆ’ 2 ๐‘ฆ ๐‘” x =^ ๐‘‰๐‘ฅ๐‘ก ๐‘ก = โˆš โˆ’ 2 ( 78. 4 ๐‘š) โˆ’ 9. 8 ๐‘š/๐‘ ^2 = 4s^ x =^5.^0 ๐‘š/๐‘ ^ (^4 ๐‘ )^ = 20ms Answer: t = 4s, x = 20m
  2. A player kicks a football from the ground level with a velocity 27.0 m/s at an angle of 30^0 above the horizontal. Find its โ€˜hang timeโ€™, horizontal distance and its maximum height. Given: Vo =27 m/s, ฮธ=30โˆ˜, g = 9. 8 ๐‘š/๐‘ ^2 Unknown: T, R=x, ๐‘Œ๐‘š๐‘Ž๐‘ฅ Solution: ๐‘‡ = 2 ๐‘ฃ๐‘œ๐‘ ๐‘–๐‘›๐œƒ ๐‘”

๐‘ฃ๐‘œ๐‘ ๐‘–๐‘›^22 ๐œƒ ๐‘”

๐‘ฃ๐‘œ^2 ๐‘ ๐‘–๐‘›^2 ๐œƒ 2 ๐‘” ๐‘‡ = 2 (^27 ๐‘ ๐‘š )๐‘ ๐‘–๐‘›30ยฐ

  1. 8 ๐‘š/๐‘ ^2 = 2.75s^ ๐‘…^ =^ ( 27 ๐‘š/๐‘ )^2 sin( 2 โˆ—30ยฐ)

  2. 8 ๐‘š/๐‘ ^2 = 64.422m^ ๐‘Œ^ =^ ( 27 ๐‘š/๐‘ )^2 ๐‘ ๐‘–๐‘›^2 30ยฐ 2 (^9. ๐‘ ^82 ๐‘š ) = 9.298m Answer: T = 2.75s, R = 64.42m, T = 9.3m

  3. A long jumper leaves the ground with an initial velocity of 10.5 m/s at an angle of 25-degrees above the horizontal. What is the time of flight, the horizontal distance, and the peak height of the long-jumper? Given: Vo = 10.5 m/s, = 25ยฐ , 2 ๐œƒ = 50ยฐ, g = 9. 8 ๐‘š/๐‘ ^2 Unknown: ๐‘ก๐‘“๐‘™๐‘–๐‘”โ„Ž๐‘ก, R=x, ๐‘Œ๐‘š๐‘Ž๐‘ฅ Solution: ๐‘… = ๐‘ฃ๐‘œ๐‘ ๐‘–๐‘›^22 ๐œƒ ๐‘”

๐‘ฅ ๐‘‰๐‘œ๐‘๐‘œ๐‘ ๐œƒ

1 2

๐‘”๐‘ก^2

( 10. 5 ๐‘š/๐‘ )^2 sin 50ยฐ

  1. 8 ๐‘š/๐‘ ^2 = 8.62m = x ๐‘ก๐‘“๐‘™๐‘–๐‘”โ„Ž๐‘ก =
  2. 62 ๐‘š
  3. 5 ๐‘š ๐‘  cos^ (25ยฐ)^ = 0.91s ๐‘Œ๐‘š๐‘Ž๐‘ฅ =
  4. 5 ๐‘š ๐‘ 

1 2 ( 9. 8 ๐‘š/๐‘ ^2 )( 0. 46 )^2 = 1.00m Answer: R = x = 8.62m, ๐‘ก๐‘“๐‘™๐‘–๐‘”โ„Ž๐‘ก = 0. 91 ๐‘  , ๐‘Œ๐‘š๐‘Ž๐‘ฅ = 1. 00 ๐‘š

  1. During World War I, the Germans constructed an enormous railway gun that was used to bombard Paris from a distance of 120 km.

(a) What was the minimum velocity necessary for the gun to have had such a range? Given: R = 120km = 120000 m Unknown: Vo Solution: ๐‘‰๐‘œ = โˆš ๐‘…๐‘” ๐‘ ๐‘–๐‘› 2 ๐œƒ or ๐‘…= ๐‘ฃ๐‘œ^2 ๐‘” sin2๐œƒ^ =^ ๐‘‰๐‘œ^ =^ โˆš๐‘…๐‘” ๐‘‰๐‘œ = โˆš 120000 ๐‘š( 9. 8 ๐‘š/๐‘ ^2 ) sin ( 2 โˆ— 45 ) = 1084.44m/s ๐‘‰๐‘œ = โˆš 120000 ๐‘š ( 9. 8 ๐‘š/๐‘ ^2 ) = 1084.44m/s Answer: Vo = 1084.44m/s (b) What was the minimum time between firing and impact when the gun was adjusted for maximum range? Solution: ๐‘ก = 2 ๐‘‰๐‘œ๐‘ ๐‘–๐‘›๐œƒ ๐‘” ๐‘ก = 2 ( 1084. 44 )๐‘š/๐‘  ๐‘ ๐‘–๐‘›45ยฐ

  1. 8 ๐‘š/๐‘ ^2 = 156.49 s