a body is thrown down from a tower of height 192 m with initial velocity 2m/s. find the time taken to hit the ground, velocity just before it hits the ground

t = 6.47 s

v = 61.38 m/s

Explanation:

List out the variables we have/are given:

Δx = -192 mv₀ = 2 m/sv = ?t = ? a = -9.8 m/s²

We can use this constant acceleration equation that contains Δx, v₀, a, and v to solve for v, the final velocity:

v² = v₀² + 2aΔx

Plug in known values:

v² = (2)² + 2(-9.8)(-192)

Simplify the equation.

v² = 4 +2(1881.6) v² = 3767.2

Take the square root of both sides.

v = 61.37752031

The velocity just before the body hits the ground is 61.38 m/s.

Let's find another equation that contains Δx, v₀, a, t to solve for t, the time:

Δx = v₀t + 1/2at²

Plug in known values:

-192 = (2)t + 1/2(-9.8)t²

Simplify this equation.

-192 = 2t - 4.9t²

Set this equation equal to 0.

0 = -4.9t² + 2t + 192

Factor this equation. You should get:

t = 6.46709390, t = -6.05893064

Since time can't be negative, we will use the positive t value.

The time taken to hit the ground is 6.47 seconds.

Is a  vector quantity  that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.

No bc a lot of things r still theories n unproven