Question

A baseball is thrown directly upward at time t= 0 and is caught again at time t=5s. Assume that air resistance is so small that it can be ignored and that the zero point of gravitational potential energy is located at the position at which the ball leaves the thrower's hand.
Sketch a graph of the kinetic energy of the baseball.

Answer 1

Answer:

Explanation:

since the ball was thrown at 0s and caought again at 5 s. applying eqaution  of motion.

$s = ut-\frac{1}{2}\times g \times t^{2}$

0 = u×5 - $\frac{1}{2}$×10×5×5

solving the eqaution we gaet initial velocity u = 25 m/s.

there fore total energy E = $\frac{1}{2}$×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                              $\rightarrow$    1

h = $ut - \frac{1}{2}\times g\times t^{2}$

applying ot in the eqaytion 1

kinetic energy = e - $mg(25t - \frac{1}{2}\times g\times t^{2})$         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.