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Sagot :
Answer:
Approximately [tex]5.42\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
Since the ball is travelling horizontally, the velocity right before landing consists of both a horizontal and a vertical component. Under the assumption that air resistance is negligible, the horizontal component would stay the same during the entire flight. Nevertheless, finding the magnitude of velocity would require finding both components.
Approach this question in the following steps:
- Find the vertical component of velocity [tex]v_{y}[/tex] right before landing using the SUVAT equation [tex]{v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}[/tex].
- Find the duration [tex]t[/tex] of the motion from the height ([tex]x_{y} = 0.750\; {\rm m}[/tex]) of the drop.
- Find the horizontal component of velocity [tex]v_{x}[/tex] by dividing horizontal displacement ([tex]s = 1.50\; {\rm m}[/tex]) by the duration [tex]t[/tex] from the previous step.
The SUVAT equation [tex]{v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}[/tex] relates the following quantities to each other:
- Vertical component of velocity right before landing, [tex]v_{y}[/tex], which needs to be found.
- Vertical component of velocity at launch, [tex]u_{y}[/tex], which would be [tex]0[/tex] assuming that the tabletop is level.
- Vertical acceleration during the flight, [tex]a_{y} = g = 9.8\; {\rm m\cdot s^{-2}}[/tex].
- Vertical displacement, [tex]x_{y} = 0.750\; {\rm m}[/tex].
Rearrange this equation to find [tex]v_{y}[/tex]:
[tex]\begin{aligned}{v_{y}} &= \sqrt{{u_{y}}^{2} + 2\, a_{y}\, x_{y}} \\ &= \sqrt{0^{2} + 2\, (9.8)\, (0.750)}\; {\rm m\cdot s^{-1}} \\ &\approx 3.83\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
To find the duration of the flight, divide the change in the vertical component of velocity [tex](v_{y} - u_{y})[/tex] by the rate of change [tex]a_{y} = 9.8\; {\rm m\cdot s^{-2}}[/tex]:
[tex]\begin{aligned}t &= \frac{v_{y} - u_{y}}{a_{y}} \\ &\approx \frac{3.83 - 0}{9.8}\; {\rm s} \\ &\approx 0.391\; {\rm s}\end{aligned}[/tex].
In other words, the duration of the flight would have been approximately [tex]0.391\; {\rm s}[/tex].
Under the assumptions, the horizontal component of velocity would have been constant during the entire flight. To find this value, divide the horizontal displacement [tex]s = 1.50\; {\rm m}[/tex] by the duration of the flight:
[tex]\begin{aligned}v_{x} &= \frac{s}{t} \approx \frac{1.50\; {\rm m}}{0.391\; {\rm s}} \approx 3.83\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Hence, the magnitude of velocity right before landing would be:
[tex]\begin{aligned}v &= \sqrt{{v_{x}}^{2} + {v_{y}}^{2}} \\ &\approx \sqrt{3.83^{2} + 3.83^{2}} \; {\rm m\cdot s^{-1}} \\ &\approx 5.42\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
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