Welcome to the article During a fireworks display a shell is shot into the air with an initial speed of tex 70 0 text m s tex at an. On this page, you will learn the essential and logical steps to better understand the topic being discussed. We hope the information provided helps you gain valuable insights and is easy to follow. Let’s begin the discussion!
Answer :
The shell explodes at a height of approximately 241.188 m, after about 6.890 s from launch, with a horizontal displacement of around 120.896 m.
To solve this problem, we can use kinematic equations for projectile motion. We'll break it down into parts:
Given:
- Initial velocity [tex](\( v_0 \))[/tex] = 70.0 m/s
- Launch angle [tex](\( \theta \))[/tex]= 75.0°
- Acceleration due to gravity [tex](\( g \)) = \( 9.81 \, \text{m/s}^2 \)[/tex]
a.Calculate the height at which the shell explodes:
To find the maximum height [tex](\( h_{\text{max}} \))[/tex] reached by the shell, we'll use the kinematic equation:
[tex]\[ h_{\text{max}} = \frac{{v_0^2 \sin^2 \theta}}{{2g}} \][/tex]
Substitute the given values:
[tex]\[ h_{\text{max}} = \frac{{(70.0 \, \text{m/s})^2 \times (\sin 75.0°)^2}}{{2 \times 9.81 \, \text{m/s}^2}} \][/tex]
[tex]\[ h_{\text{max}} \approx \frac{{4900 \times 0.965925826)}}{{19.62}} \][/tex]
[tex]\[ h_{\text{max}} \approx \frac{{4733.64015}}{{19.62}} \][/tex]
[tex]\[ h_{\text{max}} \approx 241.188 \, \text{m} \][/tex]
So, the height at which the shell explodes is approximately [tex]\( 241.188 \, \text{m} \).[/tex]
b.Calculate the time passed between the launch and the explosion:
The time taken to reach the maximum height [tex](\( t_{\text{max}} \))[/tex] can be calculated using the equation:
[tex]\[ t_{\text{max}} = \frac{{v_0 \sin \theta}}{{g}} \][/tex]
Substitute the given values:
[tex]\[ t_{\text{max}} = \frac{{70.0 \times \sin 75.0°}}{{9.81}} \][/tex]
[tex]\[ t_{\text{max}} \approx \frac{{70.0 \times 0.965925826}}{{9.81}} \][/tex]
[tex]\[ t_{\text{max}} \approx \frac{{67.61590782}}{{9.81}} \][/tex]
[tex]\[ t_{\text{max}} \approx 6.890 \, \text{s} \][/tex]
c.Calculate the horizontal displacement of the shell when it explodes:
The horizontal displacement d can be calculated using the equation:
[tex]\[ d = v_0 \cos \theta \times t_{\text{max}} \][/tex]
Substitute the given values:
[tex]\[ d = 70.0 \times \cos 75.0° \times 6.890 \][/tex]
[tex]\[ d \approx 70.0 \times 0.258819045 \times 6.890 \][/tex]
[tex]\[ d \approx 120.896 \, \text{m} \][/tex]
So, the horizontal displacement of the shell when it explodes is approximately [tex]\( 120.896 \, \text{m} \).[/tex]
Thank you for reading the article During a fireworks display a shell is shot into the air with an initial speed of tex 70 0 text m s tex at an. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
- What is the perimeter of the figure 32 0cm 51 0cm 19 0cm 63 0cm.
- The sign shows when a lift is safe to use the total mass of people should be 480 kg or less Fred and some other.
- The sign shows when a lift is safe to use the total mass of people should be 480 kg or less Fred and some other
Rewritten by : Brahmana