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:

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

Plug in known values:

Simplify the equation.

Take the square root of both sides.

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:

Plug in known values:

Simplify this equation.

Set this equation equal to 0.

Factor this equation. You should get:

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

The time taken to hit the ground is 6.47 seconds.

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

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