**1. The problem statement, all variables and given/known data**

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.)

**2. Relevant 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)

**3. The attempt at a solution**

We’re not dealing with forces, and therefor velocity stays the same through the entirety of the projection.

V_{0y}= 11sin(45°)=7.778

Solving for 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

http://ift.tt/1j18fpv