Question

Optimus decides to throw Megatron down of a building at 5.0m/s. Megatron hits the ground at 32.0m/s

- How long does it take to hit the ground?
- How far does he fall?

Answer 1

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Step-by-step explanation:

Answer 2

We have equation of motion v = u+at, where v is the final velocity, u is the initial velocity, t is the time taken and a is the acceleration.

Considering vertical motion of megatron

Here u = 0 m/s, a= 9.81
`m/s^2` and v = 32 m/s

Substituting

32 = 0+9.8*t

t = 3.27 seconds

So it will take 3.27 seconds to reach ground.

Now considering horizontal motion of megatron

We have
`s =ut+(1)/(2) at^2`, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

In this case u = 5 m/s, t = 3.27 seconds, a = 0
`m/s^2`

Substituting

`s = 5*3.27+(1)/(2) *0*3.27^2\\ \\ s = 16.35 meter`

So Megatron falls 16.35 meter away from ground.