A particle, starting from point A in the drawing at a height h0 = 3.1 m, is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height hf = 4.5 m above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A.
Well, this particle seems to be quite the little daredevil! Going down a curved runway and then shooting straight up into the air! I hope it remembered to buckle up for safety!
Now, let's get to the question at hand. To find the speed at point A, we can use the conservation of energy principle.
At point A, the particle has potential energy (mgh0) and kinetic energy (½mv^2). When it reaches point B, it has potential energy (mghf) and zero kinetic energy, as it's momentarily at rest.
Since energy is conserved, we can equate the two expressions:
mgh0 + ½mv^2 = mghf
Since mass (m) is common to both sides, we can cancel it out:
gh0 + ½v^2 = ghf
Now, we can rearrange the equation to solve for v, the speed at point A:
v^2 = 2g(hf - h0)
v = √(2g(hf - h0))
Plugging in the values, with g being the acceleration due to gravity (approximately 9.8 m/s^2), hf = 4.5 m, and h0 = 3.1 m, we can calculate the speed at point A:
v = √(2 * 9.8 * (4.5 - 3.1)) = √(2 * 9.8 * 1.4) = √(27.44) = 5.24 m/s
So, the speed of the particle at point A is approximately 5.24 m/s. That's quite a speedy little particle!
Now, let's see if it's ready for its next adventure!