Answer to Question #258239 in Physics for ziahsakdn

Question

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) ?

Answer 1

1. The maximum height is

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

2. The speed was

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

3. Range

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

Coordinates:

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