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

Step-by-step 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)

$t=\frac{12 m/s}{9.8 m/s^{2}}$ (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.