A ball with a mass of 275 g is dropped from rest, hits the floor, and rebounds upward. If the ball hits the floor with a speed o
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1) Physical principles:
a) Total mechanical energy = kinetic energy + potential energy.
b) Total mechanical energy is conserved (neglecting external forces, like drag and friction)
2) Notation:
a) Total mechanical energy: ME.
b) Kinetic energy: KE
c) Gravitational potential energy: PE
∴ ME = KE + PE = constant
3) Solution:
a) Since, ME is conserved, it is constant and would be represented in the graph by a horizontal line.
b) At start (t = 0), the ball has only KE, so KE =ME = E and PE = 0
c) As the time goes, the ball gains altitude (PE increases) and loses speed (KE decreases).
d) PE increases from 0 to a maximum value. In the graph that happens at t = 2s.
At that point, KE = 0, and PE = ME.
That is the point of highest altitude and where the speed is zero.
d) From t = 2 seg, the ball starts to lose altitude, then the ball loses PE, and gains KE.
Just before reaching the ground, at t = 4s, the ball has the same initial KE and PE as at t = 0: KE = ME and PE = 0.
The PE may be sketched on the same graph along with the KE and the ME.
The graph is attached. The red line is the ME and the blue line is the PE.
Note that at any point in the graph PE + KE = ME.
The cat fell 1.0 m from the ground.
Using the formula
Here,
Solving for t, the time it spent in the air is
The cat does not accelerate along the horizontal, so it has constant horizontal velocity. Since it strikes the floor 2.2 m from the table, then