Answer:
Explanation:
since the ball was thrown at 0s and caought again at 5 s. applying eqaution of motion.
[tex]s = ut-\frac{1}{2}\times g \times t^{2}[/tex]
0 = u×5 - [tex]\frac{1}{2}[/tex]×10×5×5
solving the eqaution we gaet initial velocity u = 25 m/s.
there fore total energy E = [tex]\frac{1}{2}[/tex]×m×25×25 J
where m is the mass of the ball according to conservation of energy E remains constant
conservation of energy:
kinetic + potential energy of the ball = E
kinetic energy = E - mgh [tex]\rightarrow[/tex] 1
h = [tex]ut - \frac{1}{2}\times g\times t^{2}[/tex]
applying ot in the eqaytion 1
kinetic energy = e - [tex]mg(25t - \frac{1}{2}\times g\times t^{2})[/tex] 2
Therefore kinetic energy vs height will be a straight line with negative slope and kinetic energy vs time will be parabola that is open upward.
