Answer:
[tex]0\; \rm m \cdot s^{-1}[/tex].
Explanation:
In a projectile motion, the kinetic energy (KE) of the projectile is converted to gravitational potential energy (GPE) and then from GPE back to KE.
In this example, the spring toy is the projectile. Refer to the diagram attached.
- The spring toy started with a velocity of [tex]12\; \rm m\cdot s^{-1}[/tex], meaning that its initial KE is non-zero.
- On the way up, the KE of this spring toy gets converted to GPE.
- At the top of the trajectory, the GPE of this toy is maximized while its KE is minimized (zero).
- As the toy returns to the ground, the GPE of this toy gets converted back to KE.
- The GPE of this toy is zero when the toy is on the ground.
In other words, the kinetic energy (KE) of this toy would be [tex]0[/tex] when it is at the top of the trajectory and GPE is maximized.
Since the KE of this toy at the top of the trajectory would be [tex]0[/tex], the velocity of this toy at that moment would also be [tex]0\![/tex].
