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?

1 Answer
Feb 20, 2017

Answer:

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

Explanation:

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.