Respuesta :
Answer:
Horizontal distance from the foot of the cliff is 40 m
Explanation:
Given :
Speed of stone [tex]v = 20\frac{m}{s}[/tex]
Vertical distance [tex]d = 20[/tex] m
Gravitational acceleration [tex]g = 10[/tex] [tex]\frac{m}{s^{2} }[/tex]
From the kinematics equation,
[tex]d = v_{o} t + \frac{1}{2} at^{2}[/tex]
Where [tex]v_{o} =[/tex] initial velocity ( [tex]v_{o} = 0[/tex] )
So time take to fall down,
[tex]t = \sqrt{\frac{2d}{a} }[/tex]
[tex]t = \sqrt{4}[/tex]
[tex]t = 2[/tex] sec
Now find horizontal distance travel by stone,
[tex]x = vt[/tex]
[tex]x = 20 \times 2[/tex]
[tex]x = 40[/tex] m
Therefore, the horizontal distance travel by stone is 40 m
The horizontal distance at which the stone strikes the ground from the foot of the cliff is 40 meters.
Given the following data:
- Displacement = 20 meters
- Horizontal speed = 20 m/s.
- Acceleration due to gravity = 10 [tex]m/s^2[/tex]
To find the horizontal distance at which the stone strikes the ground from the foot of the cliff:
First of all, we would calculate the time it took the stone to reach the ground by using the second equation of motion:
Mathematically, the second equation of motion is given by the formula;
[tex]S =ut +\frac{1}{2} at^2[/tex]
Where:
- S is the displacement.
- u is the initial velocity.
- a is the acceleration.
- t is the time measured in seconds.
Substituting the given parameters into the formula, we have;
[tex]20 = 0(t) + \frac{1}{2} 10t^2\\\\20 = 5t^2\\\\t^2=\frac{20}{5} \\\\t^2=4\\\\t=\sqrt{4}[/tex]
Time, t = 2 seconds
Now, we can find the horizontal distance:
[tex]Horizontal \;distance = horizontal \;speed \times time\\\\Horizontal \;distance =20 \times 2[/tex]
Horizontal distance = 40 meters
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