Question

The distance s (in m) above the ground for a projectile fired vertically upward with a velocity of 4[a] m/s as a function of time t (in s ) is given by s=4[a]t−4.[b]t2..

a=3, b=7


find LaTeX: t for LaTeX: v=0, find find LaTeX: v for LaTeX: t=4, find find LaTeX: v for LaTeX: t=5. What conclusions can you draw?


Find the answers to these questions. What conclusions can you draw?

Answer 1

Answers:

a) 1.224 s

b) -27.2 m/s

c) -37 m/s

Explanation:

According to the given information the distance
`s` is given by:

`s=4(3)t-4(7)t^(2)=12 t - 28t^(2)` (1)

In additio, since the projectile was fired vertically we can use the following equation:

`V=V_(o)-gt` (2)

Where:

`V` is the final velocity

`V_(o)=12 m/s` is the initial velocity

`g=9.8 m/s^(2)` is the acceleration due gravity

`t` is the time

Knowing this, let's begin:

a) Find
`t` for
`V=0`

Using (2):

`0=V_(o)-gt` (3)

`0=12 m/s-9.8 m/s^(2)t` (4)

`t=(12 m/s)/(9.8 m/s^(2))` (5)

`t=1.224 s` (6)

b) Find
`V` for
`t=4 s`

In this case we have to find
`V`:

`V=12 m/s-(9.8 m/s^(2))(4 s)` (7)

`V=-27.2 m/s` (8)

c) Find
`V` for
`t=5 s`

`V=12 m/s-(9.8 m/s^(2))(5 s)` (9)

`V=-37 m/s` (10)

Conclusions:

The time we found in a)
`t=1.224 s`when
`V=0` is at the maximum height of the projectile, just before going down. That is why in parts b) and c) when time is
`4 s` and
`5 s` the velocity is negative, because it is directed downwards.