Question #258239

A cannonball was launched at the speed of 175 m/s into the air at an angle of 60° above the ground. As it reaches maximum height, a wind with air current speed of 300 m/s sent the ball perpendicularly off-course.

 

  1. What is the maximum height?
  2. What is the cannonball’s speed as it was knocked off-course (with respect to the air current)?
  3. What is the cannonball’s resultant at coordinates (x, z) ?
1
Expert's answer
2021-10-31T18:12:37-0400

1. The maximum height is


h=(vsinθ)2/2g=1172 m.h=(v\sin\theta)^2/2g=1172\text{ m}.

2. The speed was


u=3002+(vcosθ)2=313 m/s.u=\sqrt{300^2+(v\cos\theta)^2}=313\text{ m/s}.

3. Range


R=ut=u2h/g=4841 m.R=ut=u\sqrt{2h/g}=4841\text{ m}.

Coordinates:

z=Rcosarctanvcosθ300=4647 m, x=Rsinarctanvcosθ300=1355 m.z=R\cos\arctan\frac{v\cos\theta}{300}=4647\text{ m},\\\space\\ x=R\sin\arctan\frac{v\cos\theta}{300}=1355\text{ m}.



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