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## Homework Statement

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 9.20 m/s at an angle of 19.0° below the horizontal. It strikes the ground 4.00 s later.

(a) How far horizontally from the base of the building does the ball strike the ground?

(b) Find the height from which the ball was thrown.

(c) How long does it take the ball to reach a point 10.0 m below the level of launching?

## Homework Equations

xf=xi+1/2(vxi +vxf)*t

xf=xi+vxi*t+1/2a*t

^{2}

vxf

^{2}= vxi

^{2}+2*a(xf-xi)

vf = vi + a*t

vxi= vi*cos([tex]\Theta[/tex])

vyi= vi*sin([tex]\Theta[/tex])

h= vi

^{2}s*in

^{2}([tex]\Theta[/tex])/(2g), g= gravity constant 9.8 m/s

^{2}, h= max height

R= vi

^{2}*sin(2[tex]\Theta[/tex])/g, R=max horizontal range

## The Attempt at a Solution

I know how to do part (a). This is what I did

vxi= vi*cos[tex]\Theta[/tex]

vxi= 9.2*cos(19)= 8.70 m/s

xf=xf=xi+vxi*t+1/2a*t

^{2}

xf= 0+8.7*4 +0= 34.8 m

But for part (b) and (c), im stuck, and I cannot do part (c) w/o doing part (b). Heres my attempt.

What I did bascially is finding the time vyf = 0, which indicates the time of the maximum height. Then I used the yf equation to find the height of ball at that time. Then I just got stuck. Please help asap!

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