How the velocity of vertically thrown ball varies with time?

The correct answer is Upwards positive. The velocity of a vertically thrown ball, with time, will be “up” in the positive direction. Gravity pulls the ball down, and since that is the only force acting on the ball, neglecting air resistance, the acceleration is always 9.8 m/s2 down or -9.8 m/s2.

What is the vertical velocity at the top of the path of a ball thrown vertically up and allowed to fall under free fall?

Question: A ball thrown vertically upward reaches a maximum height of 30 meters above the surface of Earth. At its maximum height, the speed of the ball is: Answer: 0 m/s. The instantaneous speed of any projectile at its maximum height is zero….Objects Launched Upward.

Variable	Value
v	0 m/s
Δy	FIND
a	-9.8 m/s2
t	0.40 s

When a ball is thrown vertically upwards with velocity keeps on decreasing what happens to its kinetic energy when it reaches the maximum height?

The kinetic energy becomes zero. According to the law of conservation of energy the kinetic energy is converted into potential energy. Potential energy is maximum at this height. When a ball is thrown vertically upwards, its velocity goes on decreasing.

What is the acceleration when a ball is thrown vertically upward?

When a ball is thrown vertically upward, it’s acceleration is -g(-9.8m/s) because it’s motion is in opposite direction to the gravitational attraction on the body, it’s velocity, v, at it’s highest point is zero while it’s initial velocity u is given by;

What is the final velocity of the ball when it stops?

After a certain time period t, the ball reaches a height beyond which it can’t move upwards anymore and stops there i.e. its velocity becomes zero at that height. And for this upward movement, the final velocity v2 is 0 because the ball has stopped at the end of this upward traversal.

What is the velocity of the ball at the highest point?

The velocity at the highest point is zero as the ball momentarily halts there before starting its downward movement. And the acceleration working on the ball at this point is the acceleration due to gravity (g) and this time it’s considered positive i.e. downwards as the ball is now ready to free fall. (ignoring air resistance)

What is the speed of the ball when it hits the ground?

Since the (vertical) velocity of the ball is given by (assuming g ≈ constant) v(t) = ds dt = v0 − gt, the speed of the object when hits the ground is the velocity at time t ∗, where t ∗ is the value of t when s(t) = sG = 0 (if your reference system is placed at s = sG = 0 for the ground).