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Answer :
To determine the final velocity of the pebble just before it hits the ground, we can use one of the equations of motion that relate initial velocity, acceleration due to gravity, height, and final velocity. The specific equation we'll use is:
[tex]\[ v_f^2 = v_i^2 + 2gh \][/tex]
where:
- [tex]\( v_f \)[/tex] is the final velocity,
- [tex]\( v_i \)[/tex] is the initial velocity,
- [tex]\( g \)[/tex] is the acceleration due to gravity,
- [tex]\( h \)[/tex] is the height.
Given the values:
- Initial velocity ([tex]\( v_i \)[/tex]) = [tex]\( 6.70 \)[/tex] m/s,
- Height ([tex]\( h \)[/tex]) = [tex]\( 38.8 \)[/tex] m,
- Acceleration due to gravity ([tex]\( g \)[/tex]) = [tex]\( 9.81 \)[/tex] m/s².
First, we calculate [tex]\( v_i^2 \)[/tex]:
[tex]\[ v_i^2 = (6.70)^2 = 44.89 \, \text{m}^2/\text{s}^2 \][/tex]
Next, we calculate [tex]\( 2gh \)[/tex]:
[tex]\[ 2gh = 2 \times 9.81 \times 38.8 \][/tex]
Calculating [tex]\( 2 \times 9.81 \)[/tex]:
[tex]\[ 2 \times 9.81 = 19.62 \][/tex]
Now, calculating [tex]\( 19.62 \times 38.8 \)[/tex]:
[tex]\[ 19.62 \times 38.8 = 761.256 \, \text{m}^2/\text{s}^2 \][/tex]
Summing up [tex]\( v_i^2 \)[/tex] and [tex]\( 2gh \)[/tex]:
[tex]\[ v_f^2 = 44.89 + 761.256 = 806.146 \, \text{m}^2/\text{s}^2 \][/tex]
To find the final velocity [tex]\( v_f \)[/tex], we take the square root of [tex]\( 806.146 \)[/tex]:
[tex]\[ v_f = \sqrt{806.146} \approx 28.3927103320553 \, \text{m/s} \][/tex]
Therefore, the final velocity of the pebble just before it hits the ground is approximately [tex]\( 28.39 \)[/tex] m/s.
[tex]\[ v_f^2 = v_i^2 + 2gh \][/tex]
where:
- [tex]\( v_f \)[/tex] is the final velocity,
- [tex]\( v_i \)[/tex] is the initial velocity,
- [tex]\( g \)[/tex] is the acceleration due to gravity,
- [tex]\( h \)[/tex] is the height.
Given the values:
- Initial velocity ([tex]\( v_i \)[/tex]) = [tex]\( 6.70 \)[/tex] m/s,
- Height ([tex]\( h \)[/tex]) = [tex]\( 38.8 \)[/tex] m,
- Acceleration due to gravity ([tex]\( g \)[/tex]) = [tex]\( 9.81 \)[/tex] m/s².
First, we calculate [tex]\( v_i^2 \)[/tex]:
[tex]\[ v_i^2 = (6.70)^2 = 44.89 \, \text{m}^2/\text{s}^2 \][/tex]
Next, we calculate [tex]\( 2gh \)[/tex]:
[tex]\[ 2gh = 2 \times 9.81 \times 38.8 \][/tex]
Calculating [tex]\( 2 \times 9.81 \)[/tex]:
[tex]\[ 2 \times 9.81 = 19.62 \][/tex]
Now, calculating [tex]\( 19.62 \times 38.8 \)[/tex]:
[tex]\[ 19.62 \times 38.8 = 761.256 \, \text{m}^2/\text{s}^2 \][/tex]
Summing up [tex]\( v_i^2 \)[/tex] and [tex]\( 2gh \)[/tex]:
[tex]\[ v_f^2 = 44.89 + 761.256 = 806.146 \, \text{m}^2/\text{s}^2 \][/tex]
To find the final velocity [tex]\( v_f \)[/tex], we take the square root of [tex]\( 806.146 \)[/tex]:
[tex]\[ v_f = \sqrt{806.146} \approx 28.3927103320553 \, \text{m/s} \][/tex]
Therefore, the final velocity of the pebble just before it hits the ground is approximately [tex]\( 28.39 \)[/tex] m/s.
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