Problem 3.27 A projectile is fired with an initial speed of 36.2 m/s at an angle

Question

Problem 3.27

A projectile is fired with an initial speed of 36.2 m/s at an angle of 43.0 above the horizontal on a long flat firing range.

Part A

Determine the maximum height reached by the projectile.

Express your answer using three significant figures and include the appropriate units.

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Part B

Determine the total time in the air.

Express your answer using three significant figures and include the appropriate units.

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Part C

Determine the total horizontal distance covered (that is, the range).

Express your answer using three significant figures and include the appropriate units.

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Part D

Determine the speed of the projectile 1.70 s after firing.

Express your answer using three significant figures and include the appropriate units.

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Problem 3.27

A projectile is fired with an initial speed of 36.2 m/s at an angle of 43.0 above the horizontal on a long flat firing range.

Part A

Determine the maximum height reached by the projectile.

Express your answer using three significant figures and include the appropriate units.

SubmitHintsMy AnswersGive UpReview Part

Part B

Determine the total time in the air.

Express your answer using three significant figures and include the appropriate units.

hmax =

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Part C

Determine the total horizontal distance covered (that is, the range).

Express your answer using three significant figures and include the appropriate units.

ttotal =

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Part D

Determine the speed of the projectile 1.70 s after firing.

Express your answer using three significant figures and include the appropriate units.

dtotal =

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v =

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Explanation / Answer

for Projectile,
Time of flight : 2*Vo* sin thetha /g
maximum height = (Vo * sin thetha)^2 /2g
Range = Vo^2 * sin (2*thetha) /g

A)
maximum height = (Vo * sin thetha)^2 /2g
= (36.2 * sin 43)^2 /(2*9.8)
= 31.1 m
Answer: 31.1 m

B)
Time of flight = 2*Vo* sin thetha /g
= 2*36.2* sin 43 /9.8
= 5.04 S

C)
Range = Vo^2 * sin (2*thetha) /g
= 36.2^2 * sin (2*43) /9.8
=133 m

D)
In vector form:
Vi = 36.2 cos 43 i + 36.2* sin 43 j
= 26.5 i + 24.7 j

a = -9.8 j

use:
Vf = Vi + a*t
= (26.5 i + 24.7 j) + (-9.8 *1.7 j)
=26.5 i + 8.04 j m/s

magnitude= sqrt (26.5^2 + 8.04^2)
= 27.7 m/s
Answer: 27.7 m/s

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