Query Questions 23

#$&*

course Phy 241

November 9 around 1pm.

023. `query 23*********************************************

Question: `qUniv. 3.48. (not in 11th edition) A ball is thrown at an unknown initial speed at angle of inclination 60 deg. It strikes a wall 18 m away at a point which is 8 m higher than thrown. What are the initial speed of the ball and the magnitude and angle of the velocity at impact?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Change in y=8m

Change in x=18m

Acceleration y=-g

Acceleration x=0

v0y=v0sin(60)=0.866v0

v0x=v0cos(60)=0.5v0

Use projectile eq:

x=1/2v0t (t=2x/v0)

y=x0+v0t-1/2gt^2

y=0.866v0(2x/v0)-1/2g(2x/v0)^2

v0=sqrt (-2g(18^2)/8-1.7(18))

v0=positive 17m/s

t=2x/v0

t=2(18)/17

t=2.11 seconds

vx=dx/dt=0.5(17)=8.5

vy=dy/dt=14.722-9.8(2.11)=-5.9

sqrt (8.5^2+5.9^2)=10.35m/s [MAGNITUDE]

tan^-1(-5.9/8.5)=-34.8 degress [ANGLE OF VELOCITY AT IMPACT]

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

`a** We know the following:

For y motion `dsy = + 8 m, ay = -g = - 9.8 m/s^2 and v0y = v0 sin(60 deg) = .87 v0.

For x motion `dsx = 18 m, ax = 0 and v0x = v0 cos(60 deg) = .5.

Assuming a coordinate system where motion starts at the origin:

The equation of motion in the x direction is thus

x = .5 v0 * t

and the equation of y motion is

y = .87 v0 t - .5 g t^2.

We know x and y at impact and we know g so we could solve these two equations simultaneously for v0 and t.

We begin by eliminating t from the two equations:

x = .5 v0 * t so

t = 2 x / v0.

Substituting this expression for t in the second equation we obtain

y = .87 v0 * (2 x / v0) - .5 g ( 2 x / v0) ^ 2. Multiplying both sides by v0^2 we obtain

v0^2 y = .87 v0^2 ( 2 x) - .5 g * 4 x^2. Bringing all the v0 terms to the left-hand side we have

v0^2 y - 1.73 v0^2 x = -2 g * x^2. Factoring v0 we have

v0^2 ( y - 1.73 x) = -2 g x^2 so that

v0 = +-sqrt(-2 g x^2 / ( y - 1.73 x) ) ) . Since we know that y = 8 m when x = 18 m we obtain

= +- sqrt( -9.8 m/s^2 * (18 m)^2 / ( 8 m - 1.73 * 18 m) ) = +-sqrt(277 m^2 / s^2) = +-16.7 m/s, approx..

We choose the positive value of v0, since the negative value would have the projectile moving 'backward' from its starting point.

Substituting this value into t = 2 x / v0 and recalling that our solution applies to the instant of impact when x = 16 m and y = 8 m we obtain

t = 2 * 18 m / (16.7 m/s) = 2.16 s.

Alternatively we can solve the system for v0 less symbolically and perhaps gain different insight into the meaning of the solution. Starting with the equations

x = .5 v0 * t and y = .87 v0 t - .5 g t^2

we see that impact occurs when x = .5 v0 t = 18 m so that t = 18 m / (.5 v0) = 36 m / v0.

At this instant of impact y = 8 m so substituting this and the t just obtained into the equation of motion for y we get

y = .87 v0 (36 m / v0 ) - .5 g (36 m / v0 )^2 = 8 m.

The equation .87 v0 (36 m / v0 ) - .5 g (36 m / v0 )^2 = 8 m is easily solved for v0, obtaining v0 = 16.7 m/s.

With this initial velocity we again confirm that t = 2.16 sec at impact.

Note that at t = 2.16 sec we get y = 14.4 m/s * 2.16 s - 4.9 m/s^2 * (2.16 s)^2 = 8 m, within roundoff error, confirming this solution.

We need the magnitude and direction of the velocity at impact. We therefore need the x and y components of the velocity at the t = 2.16 sec instant.

At this instant we have x and y velocities

vx = dx/dt = .5 v0 = 8.35 m/s and

vy = dy/dt = 14.4 m/s - 9.8 m/s^2 * t = 14.4 m/s - 9.8 m/s^2 * 2.16 s = -5.6 m/s, approx.

The velocity at impact therefore has magnitude sqrt( (8.25 m/s)^2 + (-5.6 m/s)^2 ) = 10 m/s, approx. and the angle is arctan(vy/vx) = arctan(-5.6 / 8.25) = -34 deg, approx.

At impact the object is moving at 10 m/s and at 34 deg below horizontal (i.e., it's on its way back down). **

STUDENT COMMENT

I think I have the right understanding of how to do this but I keep getting different answers than the given ones. They seem

to use many rounding errors. For example, from the given solution above I found this

vy = dy/dt = 14.4 m/s - 9.8 m/s^2 * t = 14.4 m/s - 9.8 m/s^2 * 2.16 s = -5.6 m/s, approx.

When doing this in my calculator, I get -6.7689, so I don't see how they think -5.6 is approximate.

INSTRUCTOR RESPONSE

looks like I approximated very roughly, using 9.8 * 2.16 = 20. Clearly 9.8 * 2.16 is closer to 21. Should have been more careful since the result was then used in a subtraction.

Remember, though, that my numbers are mostly the result of mental approximation. That is by design; they aren't intended to be all that accurate, though in most cases they should be reasonably close.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

Some different answers, probably just due to rounding and sig fig errors.

------------------------------------------------

self-critique rating #$&*:

&#Very good responses. Let me know if you have questions. &#

#$&*