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Sagot :
Certainly, let's go through the problem step-by-step.
### Given:
- The ball's height as a function of time is [tex]\( h(t) = -4.9 t^2 + 21 t + 1.2 \)[/tex].
### a) Time until the ball lands
We are asked to determine the time when the ball lands on the ground, which implies [tex]\( h(t) = 0 \)[/tex]:
[tex]\[ -4.9 t^2 + 21 t + 1.2 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where:
- [tex]\( a = -4.9 \)[/tex]
- [tex]\( b = 21 \)[/tex]
- [tex]\( c = 1.2 \)[/tex]
To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, compute the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 21^2 - 4(-4.9)(1.2) = 441 + 23.52 = 464.52 \][/tex]
The roots are then computed as:
[tex]\[ t = \frac{-21 \pm \sqrt{464.52}}{-9.8} \][/tex]
Since we are solving for time, we consider only the positive root:
[tex]\[ t_1 = \frac{-21 + \sqrt{464.52}}{-9.8} = 4.285714285714286 \][/tex]
Therefore, the ball will be in the air for approximately 4.29 seconds before it lands on the ground.
### b) Time when the ball is caught at 1 meter from the ground
Here, we need the time [tex]\( t \)[/tex] when the ball's height [tex]\( h(t) \)[/tex] is 1 meter:
[tex]\[ -4.9 t^2 + 21 t + 1.2 = 1 \][/tex]
Simplify this to:
[tex]\[ -4.9 t^2 + 21 t + 0.2 = 0 \][/tex]
Again, we use the quadratic formula:
[tex]\[ t = \frac{-21 \pm \sqrt{21^2 - 4(-4.9)(0.2)}}{2(-4.9)} \][/tex]
Compute the discriminant for this scenario:
[tex]\[ \Delta = 21^2 - 4(-4.9)(0.2) = 441 + 3.92 = 444.92 \][/tex]
The roots are then computed as:
[tex]\[ t = \frac{-21 \pm \sqrt{444.92}}{-9.8} \][/tex]
Since we are solving for time, we consider only the positive root:
[tex]\[ t_2 = \frac{-21 + \sqrt{444.92}}{-9.8} = 4.33281567382227 \][/tex]
Therefore, the ball will be in the air for approximately 4.33 seconds before it is caught by the fielder at 1 meter from the ground.
### c) Why do we disregard one of the roots?
In both parts a) and b), the quadratic equations have two roots. One root is positive and the other is negative.
We disregard the negative root because, in this physical context, time cannot be negative. The negative root does not have any meaningful interpretation for the passage of time from the initial event (the ball being hit) onwards.
So, the positive root represents the realistic time at which these events occur in the real world.
### Given:
- The ball's height as a function of time is [tex]\( h(t) = -4.9 t^2 + 21 t + 1.2 \)[/tex].
### a) Time until the ball lands
We are asked to determine the time when the ball lands on the ground, which implies [tex]\( h(t) = 0 \)[/tex]:
[tex]\[ -4.9 t^2 + 21 t + 1.2 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where:
- [tex]\( a = -4.9 \)[/tex]
- [tex]\( b = 21 \)[/tex]
- [tex]\( c = 1.2 \)[/tex]
To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, compute the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 21^2 - 4(-4.9)(1.2) = 441 + 23.52 = 464.52 \][/tex]
The roots are then computed as:
[tex]\[ t = \frac{-21 \pm \sqrt{464.52}}{-9.8} \][/tex]
Since we are solving for time, we consider only the positive root:
[tex]\[ t_1 = \frac{-21 + \sqrt{464.52}}{-9.8} = 4.285714285714286 \][/tex]
Therefore, the ball will be in the air for approximately 4.29 seconds before it lands on the ground.
### b) Time when the ball is caught at 1 meter from the ground
Here, we need the time [tex]\( t \)[/tex] when the ball's height [tex]\( h(t) \)[/tex] is 1 meter:
[tex]\[ -4.9 t^2 + 21 t + 1.2 = 1 \][/tex]
Simplify this to:
[tex]\[ -4.9 t^2 + 21 t + 0.2 = 0 \][/tex]
Again, we use the quadratic formula:
[tex]\[ t = \frac{-21 \pm \sqrt{21^2 - 4(-4.9)(0.2)}}{2(-4.9)} \][/tex]
Compute the discriminant for this scenario:
[tex]\[ \Delta = 21^2 - 4(-4.9)(0.2) = 441 + 3.92 = 444.92 \][/tex]
The roots are then computed as:
[tex]\[ t = \frac{-21 \pm \sqrt{444.92}}{-9.8} \][/tex]
Since we are solving for time, we consider only the positive root:
[tex]\[ t_2 = \frac{-21 + \sqrt{444.92}}{-9.8} = 4.33281567382227 \][/tex]
Therefore, the ball will be in the air for approximately 4.33 seconds before it is caught by the fielder at 1 meter from the ground.
### c) Why do we disregard one of the roots?
In both parts a) and b), the quadratic equations have two roots. One root is positive and the other is negative.
We disregard the negative root because, in this physical context, time cannot be negative. The negative root does not have any meaningful interpretation for the passage of time from the initial event (the ball being hit) onwards.
So, the positive root represents the realistic time at which these events occur in the real world.
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