Question

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 ​

Answer 1

t = 6.47 s

v = 61.38 m/s

Step-by-step explanation:

List out the variables we have/are given:

• Δx = -192 m
• v₀ = 2 m/s
• v = ?
• 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.