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Practice Problem sets with solutions in the course subject Physics for Electrical and Electronics Engineering
Typology: Assignments
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Problem Set 1: Projectile Motion Direction: Answer the following problems correctly. Show your complete solutions.
๐ฃ๐๐ ๐๐^22 ๐ ๐
๐ฃ๐^2 ๐ ๐๐^2 ๐ 2 ๐ ๐ = 2 (^27 ๐ ๐ )๐ ๐๐30ยฐ
8 ๐/๐ ^2 = 2.75s^ ๐ ^ =^ ( 27 ๐/๐ )^2 sin( 2 โ30ยฐ)
8 ๐/๐ ^2 = 64.422m^ ๐^ =^ ( 27 ๐/๐ )^2 ๐ ๐๐^2 30ยฐ 2 (^9. ๐ ^82 ๐ ) = 9.298m Answer: T = 2.75s, R = 64.42m, T = 9.3m
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 = 50ยฐ, g = 9. 8 ๐/๐ ^2 Unknown: ๐ก๐๐๐๐โ๐ก, R=x, ๐๐๐๐ฅ Solution: ๐ = ๐ฃ๐๐ ๐๐^22 ๐ ๐
๐ฅ ๐๐๐๐๐ ๐
1 2
( 10. 5 ๐/๐ )^2 sin 50ยฐ
1 2 ( 9. 8 ๐/๐ ^2 )( 0. 46 )^2 = 1.00m Answer: R = x = 8.62m, ๐ก๐๐๐๐โ๐ก = 0. 91 ๐ , ๐๐๐๐ฅ = 1. 00 ๐
(a) What was the minimum velocity necessary for the gun to have had such a range? Given: R = 120km = 120000 m Unknown: Vo Solution: ๐๐ = โ ๐ ๐ ๐ ๐๐ 2 ๐ or ๐ = ๐ฃ๐^2 ๐ sin2๐^ =^ ๐๐^ =^ โ๐ ๐ ๐๐ = โ 120000 ๐( 9. 8 ๐/๐ ^2 ) sin ( 2 โ 45 ) = 1084.44m/s ๐๐ = โ 120000 ๐ ( 9. 8 ๐/๐ ^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: ๐ก = 2 ๐๐๐ ๐๐๐ ๐ ๐ก = 2 ( 1084. 44 )๐/๐ ๐ ๐๐45ยฐ