A ball is thrown upward at an angle of 30° to the horizontal and lands on the top edge of a building is 20 m away. The top edge is 5 m above the throwing point. How fast was the ball thrown?

1 Answer
Michael
Feb 12, 2017

#sf(20color(white)(x)"m/s")#

Explanation:

MFDocs

The key to problems like this is to treat the horizontal and vertical components of the motion separately and use the fact that they share the same time of flight.

Horizontal Component

The horizontal component of the velocity is constant as it is perpendicular to g.

So we can write:

#sf(vcos30=20/t)#

Where t is the time of flight.

#:.##sf(vxx0.866=20/t)#

#sf(t=20/(0.866v)" "color(red)((1)))#

Vertical Component

We can use:

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

This becomes:

#sf(5=vsin30t-1/2"g"t^2)#

#:.##sf(5=vxx0.5xxt-1/2xx9.8xxt^2)#

#sf(5=vxx0.5xxt-4.9t^2" "color(red)((2)))#

We can substitute the value of t from #color(red)((1))# into the first part of #sf(color(red)((2))rArr)#

#:.##sf(5=(cancel(v)xx0.5xx20)/(0.866cancel(v))-4.9t^2)#

#:.##sf(5=10/0.866-4.9t^2)#

#sf(4.9t^2=11.547-5=6.547)#

#sf(4.9t^2=6.547)#

#sf(t^2=6.547/4.9=1.336)#

#sf(t=sqrt(1.336)=1.156color(white)(x)s)#

We can now put this value of t back into #sf(color(red)((1))rArr)#

#sf(20/(0.866v)=t)#

#:.##sf(0.866v=20/t=20/1.156)#

#:.##sf(v=20/(1.156xx0.866)=19.97color(white)(x)"m/s")#

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