A water balloon is thrown by Sandra out of her dormitory window at an angle of 60o. It lands at the feet of Chris, who is standing a horizontal distance of 6.00 m from the building. If it takes 2.50 s to reach Chris’s feet ?

Question

how to determine the height h that Sandra threw it from.

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Answer 1 · 2015-12-13T17:57:22

$20.2 m$ $20.2"m"$

The situation is like this:

hyperphysics.phy-astr.gsu.edu

In the convention I use, up is +ve, down is -ve.

$θ = 60^{\circ}$ $theta=60^@$

$x = 6 m$ $x=6"m"$

The horizontal component of the velocity is constant so:

$v cos 60 = \frac{x}{t}$ $vcos60=x/t$

$:.v=(x)/(tcos60)=6/(2.5xx0.5)=4.8"m/s"$

Now we can use:

$s=ut+1/2at^2$

Using the sign convention this becomes:

$h=vsinthetat-1/2"g"t^2$

$:.h=4.8xx0.866xx2.5-0.5xx9.8xx2.5^2$

$h=10.392-30.625=-20.233"m"$

This means that Sandra threw it from a height of $20.2"m"$