# Throwing a ball against a wall for projectile motion

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/s2
Time= 0=v0T-½gT2 T<0
t= (v0y+√(v0y2-2gy))/g
distance/range x=x0+V0xt
x/y velocity- v0x=Vcos(θ); v0y=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.
V0y= 11sin(45°)=7.778
Solving for t: ½(9.81)t2-7.778t-1.7 (h=1.7)
t=+1.324s

(a) I solved for total distance as regular, and subtracted (d=3.4)
v0x=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)(5.39t2)+1.7
y=4.47m

(d)
total time 1.324 – ans(b) = .784s

http://ift.tt/1j18fpv