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To determine the magnitude of the vertical velocity component of the stone just before it hits the ground, we follow several steps. Here's the step-by-step solution:
1. Initial Parameters:
- Initial speed ([tex]\( v_0 \)[/tex]) = 15 m/s
- Launch angle ([tex]\( \theta \)[/tex]) = 53° (above the horizontal)
- Height of the building ([tex]\( h \)[/tex]) = 35 meters
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.8 m/s²
2. Convert the angle to radians:
[tex]\[ \theta_{\text{rad}} = \frac{\pi}{180} \times 53° \approx 0.925 \text{ radians} \][/tex]
3. Calculate the initial vertical velocity component:
The vertical component of the initial velocity ([tex]\( v_{0_y} \)[/tex]) can be found using:
[tex]\[ v_{0_y} = v_0 \sin(\theta_{\text{rad}}) \][/tex]
Substituting the values:
[tex]\[ v_{0_y} = 15 \times \sin(0.925) \approx 11.98 \text{ m/s} \][/tex]
4. Determine the time it takes for the stone to hit the ground:
We use the kinematic equation for vertical motion:
[tex]\[ h = v_{0_y} t + \frac{1}{2} (-g) t^2 \][/tex]
Rearrange to form a standard quadratic equation in terms of [tex]\( t \)[/tex]:
[tex]\[ 0 = \frac{1}{2} g t^2 + v_{0_y} t - h \][/tex]
Substituting the values:
[tex]\[ 0 = \frac{1}{2} \times 9.8 \times t^2 + 11.98 \times t - 35 \][/tex]
Solve the quadratic equation:
[tex]\[ 4.9 t^2 + 11.98 t - 35 = 0 \][/tex]
Using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 4.9 \)[/tex], [tex]\( b = 11.98 \)[/tex], and [tex]\( c = -35 \)[/tex]:
The discriminant ([tex]\( \Delta \)[/tex]) is:
[tex]\[ \Delta = b^2 - 4ac = (11.98)^2 - 4 \times 4.9 \times (-35) \approx 816.22 \][/tex]
The positive root (since time cannot be negative) is:
[tex]\[ t = \frac{-11.98 + \sqrt{816.22}}{9.8} \approx 4.16 \text{ seconds} \][/tex]
5. Calculate the vertical velocity just before hitting the ground:
Using the kinematic equation for vertical velocity:
[tex]\[ v_{y} = v_{0_y} - g t \][/tex]
Substituting the values:
[tex]\[ v_y = 11.98 - 9.8 \times 4.16 \approx -52.76 \text{ m/s} \][/tex]
The negative sign indicates that the velocity direction is downward.
6. Magnitude of the vertical velocity component:
The magnitude of the vertical velocity component just before the stone hits the ground is:
[tex]\[ |v_y| = 52.76 \text{ m/s} \][/tex]
Therefore, the magnitude of the vertical velocity component as the rock hits the ground is approximately 52.76 m/s.
Given the choices:
a. 9.0 m/s
b. 14 m/s
c. 18 m/s
d. 26 m/s
e. 29 m/s
None of the provided options correctly match the calculated magnitude of 52.76 m/s. Hence, based on our calculations, the correct answer is not listed among the provided options.
1. Initial Parameters:
- Initial speed ([tex]\( v_0 \)[/tex]) = 15 m/s
- Launch angle ([tex]\( \theta \)[/tex]) = 53° (above the horizontal)
- Height of the building ([tex]\( h \)[/tex]) = 35 meters
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.8 m/s²
2. Convert the angle to radians:
[tex]\[ \theta_{\text{rad}} = \frac{\pi}{180} \times 53° \approx 0.925 \text{ radians} \][/tex]
3. Calculate the initial vertical velocity component:
The vertical component of the initial velocity ([tex]\( v_{0_y} \)[/tex]) can be found using:
[tex]\[ v_{0_y} = v_0 \sin(\theta_{\text{rad}}) \][/tex]
Substituting the values:
[tex]\[ v_{0_y} = 15 \times \sin(0.925) \approx 11.98 \text{ m/s} \][/tex]
4. Determine the time it takes for the stone to hit the ground:
We use the kinematic equation for vertical motion:
[tex]\[ h = v_{0_y} t + \frac{1}{2} (-g) t^2 \][/tex]
Rearrange to form a standard quadratic equation in terms of [tex]\( t \)[/tex]:
[tex]\[ 0 = \frac{1}{2} g t^2 + v_{0_y} t - h \][/tex]
Substituting the values:
[tex]\[ 0 = \frac{1}{2} \times 9.8 \times t^2 + 11.98 \times t - 35 \][/tex]
Solve the quadratic equation:
[tex]\[ 4.9 t^2 + 11.98 t - 35 = 0 \][/tex]
Using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 4.9 \)[/tex], [tex]\( b = 11.98 \)[/tex], and [tex]\( c = -35 \)[/tex]:
The discriminant ([tex]\( \Delta \)[/tex]) is:
[tex]\[ \Delta = b^2 - 4ac = (11.98)^2 - 4 \times 4.9 \times (-35) \approx 816.22 \][/tex]
The positive root (since time cannot be negative) is:
[tex]\[ t = \frac{-11.98 + \sqrt{816.22}}{9.8} \approx 4.16 \text{ seconds} \][/tex]
5. Calculate the vertical velocity just before hitting the ground:
Using the kinematic equation for vertical velocity:
[tex]\[ v_{y} = v_{0_y} - g t \][/tex]
Substituting the values:
[tex]\[ v_y = 11.98 - 9.8 \times 4.16 \approx -52.76 \text{ m/s} \][/tex]
The negative sign indicates that the velocity direction is downward.
6. Magnitude of the vertical velocity component:
The magnitude of the vertical velocity component just before the stone hits the ground is:
[tex]\[ |v_y| = 52.76 \text{ m/s} \][/tex]
Therefore, the magnitude of the vertical velocity component as the rock hits the ground is approximately 52.76 m/s.
Given the choices:
a. 9.0 m/s
b. 14 m/s
c. 18 m/s
d. 26 m/s
e. 29 m/s
None of the provided options correctly match the calculated magnitude of 52.76 m/s. Hence, based on our calculations, the correct answer is not listed among the provided options.
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