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A projectile is launched off of a 16.4 m high building with an initial velocity of 20.4 m/s at an angle of 32.1 degrees relative to the horizontal. What maximum distance in the x-direction does it travel as a projectile

A projectile is launched off of a 16.4 m high building with an initial velocity of 20.4 m/s at an angle of 32.1 degrees relative to the horizontal. What maximum distance in the x-direction does it travel as a projectile

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 A projectile is launched off of a 16.4 m high building with an initial velocity of 20.4 m/s at an angle of 32.1 degrees relative to the horizontal. What maximum distance in the x-direction does it travel as a projectile

Expert Solution
Step 1

Given data:

  • The height is H=16.4 m.
  • The initial velocity is u=20.4 m/s.
  • The angle is θ=32.1°.

The schematic diagram can be drawn as:

Advanced Physics homework question answer, step 1, image 1

The y component of the initial velocity will be:

uy=usinθuy=20.4 m/ssin32.1°uy=10.81 m/s

The acceleration in the y-direction is:

ay=-g=-9.8 m/s2

The difference in the vertical position is:

∆y=-H=-16.4 m

Now calculate the value of time as:

∆y=uyt+12ayt2-16.4 m=10.81 m/st-129.8 m/s2t24.9 m/s2t2-10.81 m/st-16.4 m=0

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