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Answer :
To solve the problem of how long Jeff's toy rocket will be in the air, we can use the physics of projectile motion. The key here is to understand that the rocket will follow a parabolic path due to the force of gravity working against it.
The equation we need to focus on includes the initial vertical velocity and the acceleration due to gravity. The standard formula for an object that is launched upward and comes back down under the influence of gravity is:
[tex]\[ h(t) = -\frac{1}{2} g t^2 + v_0 t + h_0 \][/tex]
Given:
- Initial vertical velocity [tex]\( v_0 = 25 \, \text{ft/s} \)[/tex]
- The acceleration due to gravity in feet per second squared (ft/s²) is approximately [tex]\( 32 \, \text{ft/s}^2 \)[/tex], hence the [tex]\(-\frac{1}{2} \times 32 = -16\)[/tex]
- The platform is at ground level, so [tex]\( h_0 = 0 \)[/tex]
Since we want to find out how long the toy rocket will be in the air, we need to find when it comes back to the ground. This means solving the equation for when the height [tex]\( h(t) = 0 \)[/tex].
Plugging in the values, the equation becomes:
[tex]\[ -16t^2 + 25t = 0 \][/tex]
To solve this quadratic equation, we first look for factors of the equation:
[tex]\[ t(-16t + 25) = 0 \][/tex]
This means either [tex]\( t = 0 \)[/tex] or [tex]\( -16t + 25 = 0 \)[/tex].
- [tex]\( t = 0 \)[/tex] is the initial launch time.
- Solve for [tex]\( -16t + 25 = 0 \)[/tex] for the time the rocket hits the ground:
[tex]\[ -16t + 25 = 0 \][/tex]
[tex]\[ 16t = 25 \][/tex]
[tex]\[ t = \frac{25}{16} \][/tex]
Therefore, the solution to this equation will tell us the total time the rocket is in the air. The correct equation to use based on the given options is:
[tex]\[ -16t^2 + 25t = 0 \][/tex]
This option matches the first equation provided in the list.
The equation we need to focus on includes the initial vertical velocity and the acceleration due to gravity. The standard formula for an object that is launched upward and comes back down under the influence of gravity is:
[tex]\[ h(t) = -\frac{1}{2} g t^2 + v_0 t + h_0 \][/tex]
Given:
- Initial vertical velocity [tex]\( v_0 = 25 \, \text{ft/s} \)[/tex]
- The acceleration due to gravity in feet per second squared (ft/s²) is approximately [tex]\( 32 \, \text{ft/s}^2 \)[/tex], hence the [tex]\(-\frac{1}{2} \times 32 = -16\)[/tex]
- The platform is at ground level, so [tex]\( h_0 = 0 \)[/tex]
Since we want to find out how long the toy rocket will be in the air, we need to find when it comes back to the ground. This means solving the equation for when the height [tex]\( h(t) = 0 \)[/tex].
Plugging in the values, the equation becomes:
[tex]\[ -16t^2 + 25t = 0 \][/tex]
To solve this quadratic equation, we first look for factors of the equation:
[tex]\[ t(-16t + 25) = 0 \][/tex]
This means either [tex]\( t = 0 \)[/tex] or [tex]\( -16t + 25 = 0 \)[/tex].
- [tex]\( t = 0 \)[/tex] is the initial launch time.
- Solve for [tex]\( -16t + 25 = 0 \)[/tex] for the time the rocket hits the ground:
[tex]\[ -16t + 25 = 0 \][/tex]
[tex]\[ 16t = 25 \][/tex]
[tex]\[ t = \frac{25}{16} \][/tex]
Therefore, the solution to this equation will tell us the total time the rocket is in the air. The correct equation to use based on the given options is:
[tex]\[ -16t^2 + 25t = 0 \][/tex]
This option matches the first equation provided in the list.
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