Trouble calculating a projectiles height after a certain distance

[agm1984]

trouble calculating a projectile's height after a certain distance

Homework Statement

A pebble is fired from a slingshot with a velocity of 30.0m/s. If it is fired at an angle of 30deg to the vertical, what height will it reach? If its fall is interrupted by a vertical wall 12m away, where will it hit the wall in relation to the starting position of the pebble in the slingshot?

Homework Equations

V_F² = V_o² + 2ad

and the 3 other uniform acceleration equations, i guess.

The Attempt at a Solution

ok, so i used the above equation to solve for d
0=[(30)(cos60)m/s]² + (-9.8m/s)d

d = 34.44m ~approx

my trouble comes when i try and calculate how high the pebble will be after 12 meters. if i understand correctly, the question is asking how high up the wall will the pebble hit if it is 12 meters away.

regardless, i am having difficulty creating an equation to solve for that.

it may be interesting to note that i don't know whether the ball will have passed 12m by the time it goes 34m up; however, i assume it will be beyond 12m horizontally, so the ball will still be going up when it hits the wall at 12m.

i greatly appreciate any help.

 

[haruspex]

You dropped the 2 in 2ad, but got the right value for d, so I assume that was a transcription error.
How high will the pebble be at time t? How far will it have traveled horizontally at time t?

 

[agm1984]

You dropped the 2 in 2ad, but got the right value for d, so I assume that was a transcription error.
How high will the pebble be at time t? How far will it have traveled horizontally at time t?

sorry, it was a transcription error.

i don't understand how to bring time (t) into the mix. it feels like a 2 part solution though. such as D = V*T, and use d=12m, v=horizontal velocity, and solve for t. then after that maybe do another vertical calculation for what is d after the resulting (t) seconds.

maybe something like that?

 

[haruspex]

What other kinematic equations do you know relating distance, speed and time to a constant acceleration?

 

[agm1984]

What other kinematic equations do you know relating distance, speed and time to a constant acceleration?

i suspect you are hinting towards d=V_f(t)+1/2a(t)²

 

[agm1984]

i think my confusion here is mostly in regards to translating the X and Y information.

i don't know if i am working only with vertical or if i need to utilize both X and Y to calculate this height after # seconds.

 

[haruspex]

i suspect you are hinting towards d=V_f(t)+1/2a(t)²

Yes. And for the horizontal direction?

 

[agm1984]

ok here's my plan of attack I am about to try.

im going to see if i can calculate how long it takes the projectile to travel 12 meters horizontal, and then I am going to see if i can figure out how high it will be after that amount of time.

 

[SammyS]

ok here's my plan of attack I am about to try.

im going to see if i can calculate how long it takes the projectile to travel 12 meters horizontal, and then I am going to see if i can figure out how high it will be after that amount of time.

Good plan !

 

[haruspex]

ok here's my plan of attack I am about to try.

im going to see if i can calculate how long it takes the projectile to travel 12 meters horizontal, and then I am going to see if i can figure out how high it will be after that amount of time.

That's the way.

 

[agm1984]

hmm this seems to have failed but also seems close.

i did d=v_ot +1/2at² and solved for t,
t=[2(d-v_o)]/a
t=2(12m-15m/s)/-9.8m/s²
t=0.612seconds

then i attempted to get the vertical distance after, using vertical velocity (30sin60):
d=v_ot
d=[(30)(sin60)]*(0.612)

and got d=15.91 approx

the answer in the back of the book is 17.7~ meters so I am still off. I am guessing one of the two equations i used is incorrect. preliminary suspicions aimed at d=vt probably due to gravity effects.

 

[agm1984]

hmm i got another answer using d=v_ot +1/2at² again but for vertical, and i got d=14.07m
still off

im going to try switching the equations around. d=vt for horizontal and d=vt+1/2at^2 for vertical.

 

[agm1984]

lol ok nope that failed as well, and i got vertical d=23.92m by that method.

 

[agm1984]

i think I am spinning my wheels here

 

[agm1984]

im gaining confidence that the time to go 12m is 0.8 horizontal:
d=vt
12m=15m/s*t
t=0.8

so i just need to get the right vertical distance...
a=-9.8m/s
d=?
t=0.8
v original=30sin60=25.981~
v final=?
a,t,vo=d

so it looks like i really need to use d=v_o(t) +1/2at²

but why does that give 23.9208m and not 17.7 ??

 

[kris2fer]

hmm this seems to have failed but also seems close.

i did d=v_ot +1/2at² and solved for t,
t=[2(d-v_o)]/a
t=2(12m-15m/s)/-9.8m/s²
t=0.612seconds

then i attempted to get the vertical distance after, using vertical velocity (30sin60):
d=v_ot
d=[(30)(sin60)]*(0.612)

and got d=15.91 approx

the answer in the back of the book is 17.7~ meters so I am still off. I am guessing one of the two equations i used is incorrect. preliminary suspicions aimed at d=vt probably due to gravity effects.

The 12m the question refers to is the horizontal displacement of the projectile, not the vertical one. You used the equation for vertical motion of the pebble.

 

[kris2fer]

Are you sure you're using the right velocity value in the vertical equation?

 

[agm1984]

thats right, i was thinking illogically with that attempt.

i now believe the horizontal component is simply d=v*t which solves t for 0.8seconds to reach 12m

i don't know what to do with the vertical distance though. i am frustrated now.

 

[agm1984]

The 12m the question refers to is the horizontal displacement of the projectile, not the vertical one. You used the equation for vertical motion of the pebble.

Are you sure you're using the right velocity value?

i don't see how that would be incorrect.

it is for the horizontal component. the projectile is moving horizontally 12 meters with velocity 15m/s because the actual force is 30m/s 30deg to vertical, thus horizontal force is (30cos60) = 15m/s
therefore it will take 0.8 seconds to go 12meters

 

[kris2fer]

v original=30sin60=25.981~

Where'd the 60 come from?

 

[agm1984]

the pebble is slingshotted 30 degrees to the vertical at 30meters/sec

so the resultant force is 30m/s aimed 30 deg to the vertical (aka 60 degrees to the horizontal, aka vertical force = 30sin60=25.981m/s and horizontal force = 30cos60 = 15m/s
or if you draw the triangle the other way the Y component is 30cos30 and the X is 30sin30

 

[kris2fer]

t=2(12m-15m/s)/-9.8m/s².

Sorry, my bad. I meant your velocity here is weird. How'd you get 15m/s for this if 15m/s is the velocity in horizontal direction? Also how is a metre subtracting a metre/second?

 

[agm1984]

that equation is incorrect because a=-9.8 is not true in the horizontal motion.

 

[kris2fer]

What numbers do you plug in for vertical motion?

 

[agm1984]

i don't know, i am very confused and frustrated at the moment.

the max height of the ball is v_f² = v_o² + 2ad
v_f = 0
v_o = 30sin60 or 30cos30 = 25.981m/s
a=-9.8m/s
d=unknown this is what we are solving for
t=unknown

v_f² = v_o² + 2ad
d=34.44m

after this, we want to see how high the ball will be after it travels 12m horizontally. this is where i am confused.

my current best attempt is distance=velocity*time (d=v_ot)
d= 12m
v_o= 30cos60 or 30sin30 = 15m/s
t= unknown this is what we are solving for
i am not worrying about any other variables because i don't see them relevent

d=v_ot
t=0.8s

this is where i get ultra confused because i see the answer in the back of the book is that the pebble impacts the wall at a height of 17.7m but there is no equation i can solve for d=17.7m

i feel that the answer should be d= v_ot + 0.5at²
so i am very confused what the correct method is.

 

[agm1984]

hmm i seem to have done an arithmetic error because that does net the correct answer.

d= v_ot + 0.5at²

d=(25.981)(0.8) + 0.5(-9.8)(0.8)²
d=17.6488~

and we can round that up with some sig figs (my book is all over the place but clearly they are taking 3 sig figs for 17.7m)

sorry if this thread makes anyones eyes bleed. i appreciate the help and you better believe i will dominate these questions in the future after crunching so hard on this one!

 

[haruspex]

so i just need to get the right vertical distance...
a=-9.8m/s
d=?
t=0.8
v original=30sin60=25.981~
v final=?
a,t,vo=d

so it looks like i really need to use d=v_o(t) +1/2at²

but why does that give 23.9208m and not 17.7 ??

It should give the right answer. You did d = 30sin60*0.8-9.81*0.8*0.8/2, right?

 

[kris2fer]

It should give the right answer. You did d = 30sin60*0.8-9.81*0.8*0.8/2, right?

He used +9.8 instead of -9.8.