Question

A projectile is launched with an initial velocity of 41.3 m/s at 62.2 ∘ above the horizontal. This projectile is launched from the ground level and the ground itself is level.

Part A

Calculate the magnitude of its velocity 2.05 seconds later.

Part B

What is the direction of the projectile's velocity vector 2.05 seconds after launch?

Part C

Calculate the magnitude of its velocity 5.10 seconds after launch.

Part D

What is the direction of the projectile's velocity vector 5.10 seconds after launch?

Answer 1

horizontal velocity

vx = 41.3* cos 62.2

vx = 19.262 m/s

Initial vertical velocity

uy = 41.3 sin 62.2 = 36.53 m/s

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a)

vertical velocity after 2.05 s

Vy = uy - gt

Vy = 16.443 m/s

vx = 19.262 m/s

net velocity

v^2 = vx^2 + vy^2

v = 25.326 m/s

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b)

phi = arctan ( vy/ vx) = 40.49

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c)

vy = 36.53 - 9.8* 5.1 = - 13.45 m/s

vx = 19.262 m/s

net velocity

v^2 = vx^2 + vy^2

v = 23.49 m/s

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d)

phi = ar tan ( vy/vx) = 34.925 below horizontal

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