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2 dimensional motion problem

A woman throws a ball at a vertical wall d = 3.8 m away. The ball is h = 1.9 m above ground when it leaves the woman's hand with an initial velocity of 20 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? (away from the wall)

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

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

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


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

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2 Answers

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   V0x = V0cos(theta)  and   vertical  V0y = V0sin(theta) components. 

V0x = 20cos(45)= 14.14 m/s   V0y = 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

X0=0            Xf = 3.8,  V0x = 14.4 m/s    a=0  (typical for most horizontal projection problems)
 Vfx = 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 = X0 + V0t + 1/2at2    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,   Vfy = ?

Now we need to find an equation that involves unknown we are seeking but does not involve any other unknown.  Needs to have Yf but not have Vfy.     Yf = Y0 + V0yt + 1/2at2   works

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

Now we find both Vfy and Vfx.   Vfx is easy = 14.4m/s it was constant remember.   Vfy = V0y + at      Vfy = 14.4 -9.8(0.263) = 11.82 m/s.   

Stage 2   this is as the ball turns around V0y stage 2= Vfy 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


Solving this type of problem gets more simple if one seperates the dimnsions of motion and determines which order to solve the problems in.

(b) seems easiest from first glance

V0x = 20 cos 45 = 14.14 m/s, dx = vx t, so t = dx / vx = 3.8 m / 14.14 m/s = 0.27 s

this means it takes 0.27 sec to reach the wall

(c) comes next

dy = d0y + v0y t + 0.5 ay t^2

v0y = 20 sin 45 = 14.14 m/s

dy = 1.9 + (14.14 * 0.27) - (0.5 *9.8*0.27^2) = 5.36 m

(d) comes next, but first need to find the vy that it reaches the wall at -

vy = v0y + ay t = 14.14 - (9.8 * 0.27) = 11.5 m/s

now need to use equation for dy with reference to spot ball leaves wall at -

dy = d0y + v0yt + 0.5 * ay * t^2

0 = 5.36 +11.5 t - 4.9 t^2

multiply through by -1 and rearrange

4.9 t^2 - 11.5 t - 5.36 = 0

use quadratic formula and toss out t answer that comes out negative. t = 13.45 s

(a) comes last dx = vx t = -14.14 (13.45) =  - 191 m