A ball's thrown upwards from the roof of a building with a speed of 15 m/s. Velocity of the ball when it's 5 m above the roof?

Question

6444 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-20T22:56:09

$sf(+11.27color(white)(x)"m/s")$ and $sf(-11.27color(white)(x)"m/s")$

We can use:

$sf(s=ut+1/2at^2)$

This becomes:

$sf(s=ut-1/2"g"t^2)$

$:.$ $sf(5=15t-1/2xx9.81t^2)$

$:.$ $sf(4.905t^2-15t+5=0)$

Applying the quadratic formula:

$sf(t=(15+-sqrt(225-4xx4.905xx5))/(9.81))$

This gives 2 values for $sf(t)$ :

$sf(t_1=2.677color(white)(x)s)$

and

$sf(t_2=0.38color(white)(x)s)$

These 2 times refer to the object moving up and then returning back down.

Using the convention "up is +ve" we can use:

$sf(v=u+at)$

This becomes:

$sf(v=u-"g"t)$

Using $sf(t_2rArr)$

$sf(v=15-9.81xx0.38=+11.27color(white)(x)"m/s")$

This is the velocity on the way up.

Using $sf(t_2rArr)$

$sf(v=15-9.81xx2.677=-11.27color(white)(x)"m/s")$

This is the velocity on the way down.

What goes up, must come down.