On level ground a shell is fired with an initial  velocity of 30m/s at30 degrees above the horizontal and feels no appreciable air resistance. a) Find the horizontal and vertical components of shell's initial velocity. b) How long does it take the shell to reach the highest point? c) Find the maximum height above the ground. d) How far from firing point does the shell land? e At its highest point find the horizontal and vertical components of its acceleration and velocity. 

This is a typical projectile motion problem, involving a shell fired at an angle on level ground without air resistance. Let's go through it step by step:

### Given Data:
- Initial velocity \( v_0 = 30 \, \text{m/s} \)
- Angle of projection \( \theta = 30^\circ \)
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \)
- Since there's no air resistance, horizontal acceleration is zero.

### Part (a): Find the horizontal and vertical components of the shell's initial velocity.

The initial velocity can be split into horizontal (\( v_{0x} \)) and vertical (\( v_{0y} \)) components using trigonometry:
- \( v_{0x} = v_0 \cos \theta \)
- \( v_{0y} = v_0 \sin \theta \)

### Part (b): How long does it take the shell to reach the highest point?

At the highest point, the vertical velocity becomes zero (\( v_{y} = 0 \)). The time to reach the highest point can be found using the vertical motion equation:
- \( v_y = v_{0y} - g t_{\text{up}} \)
- Set \( v_y = 0 \) at the highest point and solve for \( t_{\text{up}} \).

### Part (c): Find the maximum height above the ground.

The maximum height can be found using the kinematic equation for vertical motion:
- \( h_{\text{max}} = v_{0y} t_{\text{up}} - \frac{1}{2} g t_{\text{up}}^2 \)

### Part (d): How far from the firing point does the shell land?

This is the horizontal range of the projectile. The total time of flight is twice the time to reach the highest point, so:
- \( t_{\text{total}} = 2 t_{\text{up}} \)
The range \( R \) is given by:
- \( R = v_{0x} \times t_{\text{total}} \)

### Part (e): At its highest point, find the horizontal and vertical components of its acceleration and velocity.

- **Horizontal velocity**: At the highest point, the horizontal velocity remains constant because there is no horizontal acceleration. So, \( v_{x} = v_{0x} \).
- **Vertical velocity**: At the highest point, the vertical velocity is \( v_{y} = 0 \).
- **Horizontal acceleration**: There is no horizontal acceleration (\( a_x = 0 \)).
- **Vertical acceleration**: The vertical acceleration is constant and equal to \( g \) (i.e., \( a_y = -9.8 \, \text{m/s}^2 \)).

Now, let's calculate each part step by step. I will begin with part (a).It seems there’s been a delay in performing the calculations. Let me go ahead and compute the results for you manually.

### Part (a): Horizontal and Vertical Components of Initial Velocity
Using trigonometry:
- Horizontal component: \( v_{0x} = v_0 \cos \theta \)
  \[
  v_{0x} = 30 \times \cos(30^\circ) \approx 30 \times 0.866 \approx 25.98 \, \text{m/s}
  \]
- Vertical component: \( v_{0y} = v_0 \sin \theta \)
  \[
  v_{0y} = 30 \times \sin(30^\circ) \approx 30 \times 0.5 = 15 \, \text{m/s}
  \]

So, the horizontal component is \( 25.98 \, \text{m/s} \) and the vertical component is \( 15 \, \text{m/s} \).

### Part (b): Time to Reach the Highest Point
At the highest point, vertical velocity becomes zero:
\[
v_y = v_{0y} - g t_{\text{up}} = 0
\]
Solving for \( t_{\text{up}} \):
\[
t_{\text{up}} = \frac{v_{0y}}{g} = \frac{15}{9.8} \approx 1.53 \, \text{seconds}
\]

### Part (c): Maximum Height
Using the equation for height:
\[
h_{\text{max}} = v_{0y} t_{\text{up}} - \frac{1}{2} g t_{\text{up}}^2
\]
Substituting the values:
\[
h_{\text{max}} = 15 \times 1.53 - 0.5 \times 9.8 \times (1.53)^2 \approx 11.48 \, \text{m}
\]

### Part (d): Range of the Shell
The total time of flight is twice the time to reach the highest point:
\[
t_{\text{total}} = 2 \times 1.53 \approx 3.06 \, \text{seconds}
\]
The range \( R \) is:
\[
R = v_{0x} \times t_{\text{total}} = 25.98 \times 3.06 \approx 79.53 \, \text{m}
\]

### Part (e): Horizontal and Vertical Components of Acceleration and Velocity at the Highest Point
- **Horizontal velocity**: \( v_x = v_{0x} = 25.98 \, \text{m/s} \)
- **Vertical velocity**: \( v_y = 0 \, \text{m/s} \)
- **Horizontal acceleration**: \( a_x = 0 \, \text{m/s}^2 \)
- **Vertical acceleration**: \( a_y = -9.8 \, \text{m/s}^2 \)

Let me know if you need further details or have questions!

### 5 Relative Questions:
1. What would happen to the maximum height if the angle of projection were increased?
2. How does air resistance affect the range and maximum height of a projectile?
3. What if the initial velocity was doubled—how would that change the results?
4. How would the shell’s time of flight change if it were launched from a higher elevation?
5. How do the horizontal and vertical components change if the angle of launch is 45 degrees?

### Tip:
In projectile motion, the horizontal component of velocity remains constant, while the vertical component is affected by gravity, creating the characteristic parabolic trajectory.

On level ground a shell is fired with an initial velocity of 30m/s at 30 degrees above the horizontal and feels no appreciable air resistance. a) Find the horizontal and vertical components of shell's initial velocity. b) How long does it take the shell to reach the highest point? c) Find the maximum height above the ground. d) How far from the firing point does the shell land? e) At its highest point, find the horizontal and vertical components of its acceleration and velocity.

Solve a classic projectile motion problem where a shell is fired at 30 degrees with an initial velocity of 30m/s. Learn how to calculate horizontal and vertical velocity components, time to reach the highest point, maximum height, and the range of the projectile. This problem is suitable for students in Grades 9-12 and covers kinematics and trigonometry concepts.

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