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So according to my understanding when we thrown a ball towards ground we apply an additional amount of force which results in an acceleration greater than the one due to earth. When this ball reaches the ground, due to Newton's third law of motion, the ground experts a similar amount of force on the ball and as such it has an acceleration greater than the $g$ in the opposite direction and it bounces back.

However, when we simply release the ball, we do not add any extra force and the acceleration is constant throughout and this case should be similar to a ball being placed at the ground wherein the normal force and Earth's gravitational pull cancel each other due to which the ball should be at rest. This, however, is not the case and the ba bounces. Please help me understand why this is so.

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    $\begingroup$ A ball of Silly Putty pretty much just goes splat and stays there, whether thrown or not. $\endgroup$
    – Jon Custer
    Commented Jun 10 at 18:22
  • $\begingroup$ @JonCuster well there are different velocities which give different results for silly putty given its non-newtonian nature. $\endgroup$
    – Triatticus
    Commented Jun 10 at 18:24
  • $\begingroup$ @JonCuster youtu.be/A1no81m2lZE $\endgroup$
    – Solomon Slow
    Commented Jun 10 at 19:12
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    $\begingroup$ You have confused Silly Putty for Play-Doh. Silly Putty balls bounce. It's a weird substance. $\endgroup$
    – rob ♦
    Commented Jun 10 at 19:20
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    $\begingroup$ When the ball hits the ground, how does the ground know whether it was thrown from one height or dropped from a higher one? $\endgroup$
    – WillO
    Commented Jun 10 at 19:35

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When the ball reaches the ground with some momentum, it becomes compressed due to inertia (think of a spring for example). Its kinetic energy is then converted to potential energy (spring-like energy) which is then converted back into kinetic energy (it extends back into its original shape), thereby "pushing" itself up from the ground.

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when we thrown a ball towards ground we apply an additional amount of force...When this ball reaches the ground,...the ground experts a similar amount of force on the ball.

That's not quite right. Your hand can only apply force to the ball while your hand is touching it. From the moment when it leaves your finger tips until the moment when it touches the ground, the only force* acting on the ball—accelerating the ball—will be gravity.

The only difference between throwing the ball toward the ground and simply allowing it to fall naturally is the velocity. If you simply release the ball from height $h$ meters, and you let it fall, then it will hit the ground at $\sqrt{9.8\cdot{}2h}$ meters per second. If you throw it downward with initial velocity $v$ meters per second, then it will hit the ground at $v+\sqrt{9.8\cdot{}2h}$ meters per second.

The reason why it bounces back up in either case is explained by Jesse's answer: The ball acts like a spring that absorbs the kinetic energy of the collision with the ground, and then gives some of it back† by launching the ball up into the air again.

*As is customary in all elementary kinematics problems, we shall say, "...ignoring air resistance."

†A perfectly elastic ball would give all the kinetic energy back, and in the case where you simply released it, it would return to the original height from which you dropped it. But in realty, no actual bouncy ball is perfectly elastic.

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  • $\begingroup$ Note: the heuristic for whether you can ignore air resistance is whether the mass of air displaced is comparable to the mass of the object. For a sphere with roughly the density of water, the order of magnitude for nonnegligible air resistance is a travel distance $\sim10^3$ times the radius-or-diameter. $\endgroup$
    – rob ♦
    Commented Jun 10 at 19:23

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