A projectile is launched off of a 17.5 m high building with an initial velocity of 25.8 m/s at an angle of 31.6 degrees relative to the horizontal. How long is it in the air for?

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Chapter1: Units, Trigonometry. And Vectors
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17.5 m
31.6
25.8 2
Transcribed Image Text:17.5 m 31.6 25.8 2
A projectile is launched off of a 17.5 m high building with an initial velocity of 25.8 m/s at an angle of 31.6
degrees relative to the horizontal. How long is it in the air for?
Transcribed Image Text:A projectile is launched off of a 17.5 m high building with an initial velocity of 25.8 m/s at an angle of 31.6 degrees relative to the horizontal. How long is it in the air for?
Expert Solution
Step 1

Height of the building = 17.5 m, let it be the zero of our calculation (to measure the vertical distance).

Initial Velocity = 25.8 m/s at an angle of 31.6o relative to the horizontal.

Horizontal component of the velocity = 25.8 Cos( 31.6o), horizontal acceleration 0 m/s2

Vertical  component of the velocity = 25.8 Sin( 31.6o), vertical acceleration = -g m/s2 = -10 m/s2 (say)

How long does the projectile stays in air ? It will stay i air till the vertical distance becomes -17.5 m

We are not concerned with the horizontal motion as it would just move the projectile in a uniform motion in the horizontal direction, still its hits the ground.

Vertical distance to be covered = -17.5 m

Vertical velocity = 25.8 Sin( 31.6o) = 13.51884 m/s

Vertical acceleration = -10 m/s2

s= ut + 12at2 -17.5 = 13.51884t -5t2 5t2 -13.51884t-17.5=0t= 13.51884±23.081572510=3.66 or -0.95627335 s

Negative time solution is not physical 

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