Well to answer this odd question we must analyze what do we need to find. Since the final horizontal velocity is reversed we need to find this final velocity first. Projectile problems usually require you to turn the velocity at an angle into its horizontal V_{0x} = V_{0}cos(theta) and vertical V_{0y} = V_{0}sin(theta) components.

V_{0x} = 20cos(45)= 14.14 m/s V_{0y} = 20sin(45) = 14.14m/s

Now what I find easiest to do and will really help you in all kinematic problems is to do an X-Vat which stands for X position, V velocity, a acceleration, t time

So there are two stages to this problem

Stage 1 from the hand to wall

X_{0}=0 X_{f} = 3.8, V_{0x} = 14.4 m/s a=0 (typical for most horizontal projection problems)

V_{fx} = 14.4m/s (reason is a=0 implies velocity doesn't change) t = time ? (time is variable that is common to both horiz and vertical) That means the time it takes to cross 3.8m is same as time it takes to climb to where it will it the which is not necessarily 1.9 m.

X = X_{0} + V_{0}t + 1/2at^{2} Now when both a= 0 and X0 = 0 this just turns into Dist = Rate *time (handy to know on tests when you are struggling for correct formula. make sure only use if a=0 though)

3.8 = 14.4t So time to wall is t= 0.263 sec. Notice also that sometimes we answer problems as we go along and not necessarily in order, in this case t= 0.263 seconds answer part b.)

Now let us do a YVat y0 = 1.9 yf = ? (what looking for part c) V0y= 14.4m/s a = -9.8m/s^2 (acceleration in y direction will always be this value. some instructors use positive value for gravity here but I have found in kinematics for most part keeping a neg and paying careful attention to your signs will avoid problems) t= 0.263sec, V_{fy} = ?

Now we need to find an equation that involves unknown we are seeking but does not involve any other unknown. Needs to have Y_{f} but not have V_{fy}. Y_{f} = Y_{0} + V_{0y}t + 1/2at^{2} works

Yf = 0 + 14.4(0.263) + 1/2(-9.8)(0.263) = 2.5m answer C

Now we find both V_{fy} and V_{fx}. Vfx is easy = 14.4m/s it was constant remember. V_{fy} = V_{0y} + at Vfy = 14.4 -9.8(0.263) = 11.82 m/s.

Stage 2 this is as the ball turns around V_{0y} stage 2= V_{fy} stage 1=11.82m/s

V0x stage1 = -Vfx stage 1= - 14.4m/s Now do this process in reverse

This time flight time will be determined by vertical motion or Y-vat

Y0 = 2.5 m Yf = 0 (helpful to always treat the ground as zero) V0y = 11.82m/s a = -9.8 t=?

Vfy = ? Use the last equation we used before to finish this answer. I don't want to do more because I will be doing your homework and that is not good

## Comments

hi ben -

re - checked this answer using a graphing calculator in parametric mode -

answer (d) turns out to be 2.75 sec coming from graphs of parametric equations

x1t = -14.14 t

y1t = 5.36 + 11.5t - 4.9t^2

and solving for y1t = 0 m

rechecked using quadratic formula and guess i made an error, yes find t = 2.75 s that way as well.

this means that (a) = dx = -14.14 *2.75 = - 38.9 m