Answer :
Answer:
Approximately [tex]51\; {\rm m\cdot s^{-1}}[/tex], assuming that the air resistance on the stone is negligible, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
Under the assumption that air resistance on the stone is negligible:
- horizontal component of the velocity of the stone will be constant, while
- vertical component of velocity of the stone will change at a constant of [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex].
From the perspective of an observer on the ground, the horizontal and vertical component of initial velocity of the stone would be:
- Horizontal component: [tex]u_{x} = 15\; {\rm m\cdot s^{-1}}[/tex], and
- Vertical component: [tex]u_{y} = (-20)\; {\rm m\cdot s^{-1}}[/tex] (same as that of the balloon, negative because the motion points downward.)
After [tex]t = 5\; {\rm s}[/tex], from the perspective of an observer on the ground, the horizontal and vertical component of the velocity of the stone would be:
- Horizontal component:
[tex]v_{x}(\text{stone}) = u_{x} = 15\; {\rm m\cdot s^{-1}}[/tex]. - Vertical component:
[tex]\begin{aligned}v_{y}(\text{stone}) &= u_{y} + a\, t \\ &= (-20)\; {\rm m\cdot s^{-1}} + ((-9.81)\; {\rm m\cdot s^{-2}})\, (5\; {\rm s}) \\ &=(-69.05)\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
From the perspective of the observer on the ground, the velocity of the balloon would be:
- Horizontal component: [tex]v_{x}(\text{balloon}) = 0\; {\rm m\cdot s^{-1}}[/tex].
- Vertical component: [tex]v_{y}(\text{balloon}) = (-20)\; {\rm m\cdot s^{-1}}[/tex].
To find the velocity of the stone from the perspective of an observer in the balloon, subtract the velocity of the ballon relative to the ground from the velocity of the stone relative to the ground:
- Horizontal component:
[tex]v_{x}(\text{stone}) - v_{x}(\text{balloon}) = 15\; {\rm m\cdot s^{-1}}[/tex]. - Vertical component:
[tex]\begin{aligned} & v_{y}(\text{stone}) - v_{y}(\text{balloon}) \\ =\; & (-69.05)\; {\rm m\cdot s^{-1}} - (-20)\; {\rm m\cdot s^{-1}} \\ =\; & (-49.05)\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
The magnitude of this velocity would be:
[tex]\sqrt{(15)^{2} + (-49.05)^{2}}\; {\rm m\cdot s^{-1}} \approx 51\; {\rm m\cdot s^{-1}}[/tex].
Final answer:
The velocity of the stone right as it hits the ground as measured from an observer in the balloon is 25 m/s.
Explanation:
To find the velocity of the stone right as it hits the ground as measured from an observer in the balloon, we can use the concept of relative velocity.
First, we need to find the velocity of the stone relative to the balloon. Since the stone is thrown perpendicular to the path of descent, its horizontal velocity remains constant at 15 m/s. The vertical velocity of the stone is the same as the velocity of the balloon, which is descending with a constant velocity of 20 m/s downward.
Using the Pythagorean theorem, we can find the magnitude of the velocity of the stone as it hits the ground:
velocity = sqrt((horizontal velocity)^2 + (vertical velocity)^2) = sqrt((15 m/s)^2 + (20 m/s)^2) = sqrt(225 + 400) = sqrt(625) = 25 m/s
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