A human cannon ball is fired at an angle of 60 degrees and speed of 25m/s. If the net is located 35 meters in front of the human cannon, how high in the air should the net be placed so the human lands in the center of it. Please Use the equations provided ## Equations### Vector Components, Standard Cartesian Plane:- $ V_x = V \cos \theta $- $ V_y = V \sin \theta $- $ V_x^2 + V_y^2 = V^2 $- $ \tan \theta = V_y / V_x $### Constant Linear Acceleration:- $ v = v_0 + at $- $ d = d_0 + v_0t + \frac{1}{2} at^2 $- $ v^2 = v_0^2 + 2a(d - d_0) $### Projectile Motion:- $ x = x_0 + v_{0x}t $- $ y = y_0 + v_{0y} - \frac{1}{2} gt^2 $- $ v_y = v_{0y} - gt $### Rotational Motion:- $ 360^\circ = 2\pi $ radians- $ \theta = s/r $- $ \omega = v/r $- $ \alpha = a_T/r $- $ a_c = v^2/r = r\omega^2 $- $ \omega = \omega_0 + \alpha t $- $ \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 $- $ \omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0) $### Newton's Second Law of Motion:- $ \sum \vec{F} = m \vec{a} $### Gravitational Force, or Weight:- $ F_g = mg $### Frictional Force:- $ F_f = \mu F_n $### Centripetal Force:- $ F_c = mv^2/r $### Note:- Use $ g = 10.0 \, \text{m/s}^2 $ for the acceleration due to gravity near the surface of the Earth.

Name: A human cannon ball is fired at an angle of 60 degrees and speed of 25
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Description: A human cannon ball is fired at an angle of 60 degrees and speed of 25m/s. If the net is located 35 meters in front of the human cannon, how high in the air should the net be placed so the human lands in the center of it. Please Use the equations pro

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College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A human cannon ball is fired at an angle of 60 degrees and speed of 25m/s. If the net is located 35 meters in front of the human cannon, how high in the air should the net be placed so the human lands in the center of it. Please Use the equations provided

## Equations

### Vector Components, Standard Cartesian Plane:
- $ V_x = V \cos \theta $
- $ V_y = V \sin \theta $
- $ V_x^2 + V_y^2 = V^2 $
- $ \tan \theta = V_y / V_x $

### Constant Linear Acceleration:
- $ v = v_0 + at $
- $ d = d_0 + v_0t + \frac{1}{2} at^2 $
- $ v^2 = v_0^2 + 2a(d - d_0) $

### Projectile Motion:
- $ x = x_0 + v_{0x}t $
- $ y = y_0 + v_{0y} - \frac{1}{2} gt^2 $
- $ v_y = v_{0y} - gt $

### Rotational Motion:
- $ 360^\circ = 2\pi $ radians
- $ \theta = s/r $
- $ \omega = v/r $
- $ \alpha = a_T/r $
- $ a_c = v^2/r = r\omega^2 $
- $ \omega = \omega_0 + \alpha t $
- $ \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 $
- $ \omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0) $

### Newton's Second Law of Motion:
- $ \sum \vec{F} = m \vec{a} $

### Gravitational Force, or Weight:
- $ F_g = mg $

### Frictional Force:
- $ F_f = \mu F_n $

### Centripetal Force:
- $ F_c = mv^2/r $

### Note:
- Use $ g = 10.0 \, \text{m/s}^2 $ for the acceleration due to gravity near the surface of the Earth.

Expert Solution

Step 1

Given,

A human cannon ball is fired at an angle of 60 degrees and speed of 25m/s.

So,

$v_{0 x} = (25 m / s) \cos 60 ° = 12.5 m / s v_{0 y} = (25 m / s) \sin 60 ° = 21.65 m / s$

If the net is located 35 meters in front of the human cannon , so x = 35 m and we need to calculate value of y.

x = $x_{0} + v_{0 x} t, x_{0} = 0, s o x = v_{0 x} t$

Hence t = $\frac{x}{v_{0 x}} = \frac{35 m}{12.5 m / s} = 2.8 s$

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