Calculate to what height a ball rises?

Question

A ball is shot from the ground into the air. At a height of 9.1m, its velocity is (7.6i+6.1j)m/s. (i horizontal and j upward)
a) to what max. height does the ball rise?
b) what horizontal distance does the ball travel
(X-Xo=(Vo.COSangle)t right?)
c) what are the magitude and d) angle (below the horizontal) of the ball's velocity just before it hits the ground?


I did figure some parts out; since i is horiz. and j is upward the ball is still climbing. The magnitude of v is sq root of (7.6^2)+(6.1^2)=9.75 m/s and the angle at that moment is arctan(6.1/7.6) = 38.75degr.

I thought by calculating the horizontal range R I would get somewhere but I'm stuck again.
I think with the instantaneous velocity after 9.1m I must be able to calculate the t and the Vo and go from there but how?
Who can help me?

Answer 1

ok lets start with the generic approach to physics problems...first pick a coordinate system. for this problem i will let the ground (where the ball is shot from) be the origin...list everything you know and all the equations corresponding to what you know. paticularly at 9.1m we can use the i and j components of the vector and our equations to find the time it occured (*note* in my notation Vo Xo To are all initial conditions and Vxt1 is the i component of the velocity at time T1....Vxt2 is the j component of the velocity at T2, To for this problem (our initial time) will be zero seconds)


what we dont know
--------------------------
To=?
T1=when the ball is at 9.1M
Xf=the distance from Xo that the ball travels before hitting the ground
Vf=final velocity
Vfx=final i component of the velocity
etc etc

what we know
---------------------
Xo=0
Yo=0
Ax=accel. in the x direction = 0
Ay= -g = -9.8 m/s^2
Vxt1=7.6 m/s
Vyt1=6.1m/s



ok so it seems we really dont know much at all! but lets look at our equations before we despair.

Vx=Vox+Ax*Tf if we plug in only our Ax = 0 we get Vx = Vox and this makes sense if we consider that there is NO accel in the x direction so our Vx at any time will be the same. But we really need to prove this w/ our formulas so lets do just that. lets focus in on our problem so that the beginning is when it leaves the ground and the end is when it reaches 9.1m high. at this time is what we will (at this point) call Tf

Vx=Vox + Ax*t and Ax is zero so we have Vx=Vox...and our Vx is 7.6 m/s so our Vox must be 7.6m/s

Vox=7.6m/s

X=Xo+Vox*t + 1/2*Ax*tf^2...well Ax is zero and our Xo is 0 and Vox = 7.6m/s so what we get is

X=7.6*t.....well this isnt useful..we dont know x or t....lets see if we can solve for t. we've used our x equations so lets see what we can glean from our y ones...

Vy = Voy + Ay*Tf....well Ay is -9.8....Vy is 6.1 m/s so if we plug in we get

6.1=Voy -9.8*t......well we still have one equation and 2 unknowns...so lets use another y equation

Y=Yo + Voy*t + 1/2*Ay*Tf^2 we know Yo is 0...AND we know that Y is 9.1.....we also know Ay is -9.8....substituting in we get

9.1=0 + Voy*t -4.9*t^2....well finally...we have two equations and two unknowns....


9.1=0 + Voy*t -4.9*t^2
and
6.1=Voy -9.8*t

now this will get messy but bear with me.
lets solve the second for Voy...
6.1 = Voy -9.8*t
+9.8*t +9.8t

6.1+9.8t = Voy...plug this into our first equation

9.1=(6.1+9.8t )*t -4.9*t^2
and lets do some algebra and simplify that into
9.1=6.1*t + 9.8*t^2 - 4.9*t^2...simplify further and you get

0 = 4.9*t^2 +6.1*t -9.1....now you use the quadratic formula on this beast and you get t= 0.876seconds and t = -2.121 seconds...well time cant be negative so we do away with the second and we know that the ball rises to 9.1m in 0.876 seconds

now remember our other equations?

6.1=Voy -9.8*t
and
Vox=7.6m/s

well lets get our Voy since we know that that equation is true when our t is 0.876 seconds...when we plug that in we get

Voy = 14.685 approximately

ok now we only keep our Voy and our Vox since we are going to completely redo what we define as Tf .....this time lets work the problem (now knowing Voy and Vox) where Tf is going to be when it hits the ground...and that will be the correct clock to have to awnser the questions given....lets start by just getting our equations and plugging stuff in and finding out what the ball is doing

X=Xo + Vox*t + 1/2 Ax*t^2....well Xo is 0....and Vox is 7.6 m/s and Ax is 0...plugging in we get

X=7.6*t......we now need to find the t when the ball hits the ground so we can find how far it traveled and awnser question b.....so lets plug stuff into our y equations

Y=yo + voy*t + 1/2 *a *t^2....well our Y is when the ball hits the ground so it will be 0....yo is 0......voy is 14.685 m/s and a for this is -9.8m/s^2...plugging in we get

0=0+14.685*t-4.9*t^2....great we can now get our t (which is the total time the ball was in the air)..we just factor out a t and get
t*(14.685-4.9*t)=0 well now we set both to 0 and get t=0 and t = 2.997....well this makes sense since the ball is at Y=0 when it is initally shot and when it hits the ground...so our Tf is 2.997s...lets plug this into

X=7.6*t and we get X is 22.8m...and bam we have the awnser to b......now lets get part a....when the ball is at its maximum height...that is cake now that we have all our core parts to our equations.....when it is at its apex its velocity in the y direction has to be 0 because before that its positive...then after its negative so it has to cross 0...so Vy will be 0....we know that in 2-d projectile motion it takes just as much time to go from initial to apex as it does from apex to ground....so our t for this will be our final t divided by 2...whichwill be 1.498s....plugging our new numbers into

Y=Yo + Voy*t + 1/2 * a * t^2 we get

Y=0 + 14.685*1.498 + 1/2 * (-9.8) * (1.498)^2....do some quick number crunching and our Y is 11.....nice...we have the awnser to a....11meters...now for the final question!!! we need the final velocity's magnitude and its direction....bah cake!

Vy = Voy + Ay*t.....

Vy=14.685 -9.8*2.997....our Vy is -14.686 m/s

Vx=Vox + Ax*t....
Vx=7.6+0*2.997....Vx = 7.6m/s.....well we know that Vx and Vy makes a right triangle...so the magnitude of it is just using the Pythag's theorm.....sqrt(7.6^2 + (-14.686)^2)...we get bum bum bum!!!!! 16.536 m/s....ok now for the angle....well still on the subject of right triangles...we know that the tangent of the angle is Y/X.....so the angle is arctan(y/x)....apply this to our right triangle we get

angle = atan(-14.686/7.6)...which is!!!!!!! -62.638 degrees!!!


now my arithmatic SUCKS....so im suprised if i got this problem completely right w/o making any mistakes just typing this all out...so work through my explanation.....make sure i didnt make any arithmatic mistakes and you'll have your awnsers and hopefully a better understanding of how to work these problems....just remember....list what you know.....list your equations....plug in and find EVERYTHING you can...then read the question...by then you'll already know everything about what is going on in the problem and wont have any problems awnsering it....=D have a great day

Answer 2

Assuming you're not accounting for air friction, and level ground:

You did a nice job of calculating the current speed and angle, but both are useless to finding your answers.

Horizontal velocity will always be the same.

You have the vertical componet of the velocity and the altitude.
You know the rate at which gravitry will be decreasing it's vertical speed.
How much velocity did it lose in it's 9.1m rise, and how long did it take. You can figure out the initial vertical velocity by that.

From the initial vertical velocity you can figure out the height when the vertical velocity will be zero, and the time.

The range will be twice that time multiplied by the initial (constant) horizontal velocity.

The angle it was fired at can be calculated by the INVTAN of the initial horiz and vertical velocities.

since you're not accounting for air friction, the angle and speed at impact will be idntical.

Answer 3

I had asked the same question several times, and did not get an answer

Answer 4

Not totally sure about this