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Psychros

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## Homework Statement

A woman throws a ball at a vertical wall d = 3.4 m away. The ball is h = 1.7 m above ground when it leaves the woman's hand with an initial velocity of 11 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component remains unchanged. (Ignore any effects due to air resistance.)

(a) Where does the ball hit the ground? meters (from the wall)

(b) How long was the ball in the air before it hit the wall? seconds

(c) Where did the ball hit the wall? meters (above the ground)

(d) How long was the ball in the air after it left the wall? seconds

## Homework Equations

gravity=-9.8m/s

^{2}

Time= 0=v

_{0}T-½gT

^{2}T<0

t= (v

_{0y}+√(v

_{0y}

^{2}-2gy))/g

distance/range x=x

_{0}+V

_{0x}t

x/y velocity- v

_{0x}=Vcos(θ); v

_{0y}=vsin(θ)

## The Attempt at a Solution

I did this for a practice problem with different values, and got the correct answers. With these values it says I've gotten all incorrect answers… If someone could please help and verify what I've done wrong!

*I made the person's hand the origin

We're not dealing with forces, and therefor we don't lose velocity due to impact.

V

_{0y}= 11sin(45°)=7.778

Solving for total time t: ½(9.81)t

^{2}-7.778t-1.7 (h=1.7)

t=+1.324s

(a) I solved for total distance as regular, and subtracted (d=3.4)

v

_{0x}=11cos(45)=7.778

X=0+7.778(1.324)-3.4

x=10.29-3.4= 6.898m

(b) i made the x value 3.4 (distance)

3.4=0+7.778t

t=.539s

(c ) I plugged in the time it took to get to the wall to the vertical formula, and added h 1.7

y=7.778(5.39)-½(9.81)(5.39t

^{2})+1.7

y=4.47m

(d)

total time 1.324 - ans(b) = .784s

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