a tight rope walker drops a coin from a height of 3150m above the ground how high is it when it is traveling at 60m/s
neglecting air resistance...
Vf^2=Vi^2+2ad
a=-9.8 d= 3150-h
solve for h.
10
To find the height of the coin when it is traveling at a certain speed, you need to consider the motion of the coin and apply the laws of physics. In this case, we can use the equations of motion to solve for the height.
First, let's consider the initial height of the coin, which is 3150m. We'll label this as h₀.
Next, we'll assume that the only force acting on the coin is gravity, which causes it to accelerate downwards at a rate of 9.8 m/s². We'll label the initial velocity of the coin as v₀, which is zero since it's dropped from rest.
The equation we can use to solve for the height of the coin at a given velocity is:
h = h₀ + (v₀²/2g)
where:
h is the final height
h₀ is the initial height
v₀ is the initial velocity
g is the acceleration due to gravity (approximately 9.8 m/s²)
In this case, the initial velocity is zero since the coin is dropped, so the equation simplifies to:
h = h₀ + (0²/2g)
= h₀
Therefore, the height of the coin when it is traveling at 60 m/s is still 3150 meters above the ground.