Question

A small first-aid kit is dropped by a rock climber who is descending steadily at 1.9 m/s. After 2.4 s, what is the velocity of the first-aid kit? The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s.

Answer 1

Answer:

$v=-21.65 m/s$

Explanation:

From the exercise we have:

$v_{o}=1.9m/s\\ g=9.81 m/s^{2}\\ t=2.4s$

To find the velocity after 2.4s we need to use the following formula:

$v=v_{o}+gt$

$v=1.9m/s-(9.81m/s^{2})(2.4s)=-21.65m/s$

The negative sign means that the kit is going down.

Answer 2

Answer:

$v_f=25.444\ m/s$

Explanation:

All given this are:

Initial velocity is , $v_i=1.9\ m/s$

Time taken, $t=2.4 \ s$

The acceleration due to gravity is , $g=9.81 \ m/s^2$

So , we need to find final velocity, $v_f$.

So we will use equation of motion.

$v_f-v_i=a\times t$. Here a is acceleration which is g .

putting all those values.

$v_f=1.9 + 9.81\times2.4=25.444\ m/s$.

Hence , it is the required solution.