# Trouble calculating a projectiles height after a certain distance

• agm1984
In summary, the problem involves finding the height of a projectile fired from a slingshot with a velocity of 30.0m/s and an angle of 30deg to the vertical. The projectile reaches a height of 34m and then travels 12m horizontally before hitting a vertical wall. The goal is to calculate the height of the projectile at the point of impact with the wall. To do this, the equations VF2 = Vo2 + 2ad and d=Vf(t)+1/2a(t)2 are used to find the time it takes for the projectile to travel 12m horizontally and the corresponding vertical distance after that time. However, the initial velocity used in the vertical equation may be incorrect, leading
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

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.

Last edited:
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?

haruspex said:
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?

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

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

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

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.

agm1984 said:
i suspect you are hinting towards d=Vf(t)+1/2a(t)2
Yes. And for the horizontal direction?

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.

agm1984 said:
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 !

agm1984 said:
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.

hmm this seems to have failed but also seems close.

i did d=vot +1/2at2 and solved for t,
t=[2(d-vo)]/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=vot
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.

Last edited:
hmm i got another answer using d=vot +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.

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

i think I am spinning my wheels here

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=vo(t) +1/2at2

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

agm1984 said:
hmm this seems to have failed but also seems close.

i did d=vot +1/2at2 and solved for t,
t=[2(d-vo)]/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=vot
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.

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

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.

kris2fer said:
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 said:
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

agm1984 said:
v original=30sin60=25.981~
Where'd the 60 come from?

Last edited:
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

agm1984 said:
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?

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

What numbers do you plug in for vertical motion?

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

the max height of the ball is vf2 = vo2 + 2ad
vf = 0
vo = 30sin60 or 30cos30 = 25.981m/s
a=-9.8m/s
d=unknown this is what we are solving for
t=unknown

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=vot)
d= 12m
vo= 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=vot
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= vot + 0.5at2
so i am very confused what the correct method is.

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

d= vot + 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!

agm1984 said:
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=vo(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?

haruspex said:
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.

## 1. How can I calculate the height of a projectile at a specific distance?

To calculate the height of a projectile at a specific distance, you will need to use the projectile motion equations. These equations take into account the initial velocity, angle of launch, and acceleration due to gravity. By plugging in the known values, you can solve for the height at the desired distance.

## 2. What is the formula for calculating the height of a projectile at a certain distance?

The formula for calculating the height of a projectile at a certain distance is h = h0 + v0y * t - 1/2 * g * t^2, where h0 is the initial height, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time at the desired distance.

## 3. Can I use the same formula for calculating the height of a projectile at any distance?

Yes, the formula for calculating the height of a projectile at a certain distance can be used for any distance as long as the initial conditions remain the same. This means that the initial velocity, angle of launch, and acceleration due to gravity must remain constant throughout the entire trajectory of the projectile.

## 4. Is there a specific unit of measurement for the height of a projectile?

The height of a projectile can be measured in any unit of length, such as meters or feet. It is important to be consistent with the units used for the initial conditions and the distance at which the height is being calculated.

## 5. Can I calculate the height of a projectile without knowing the initial velocity and angle of launch?

No, the initial velocity and angle of launch are necessary for calculating the height of a projectile. Without these values, the projectile motion equations cannot be used to determine the height at a certain distance. These values can be measured or provided in a problem statement.

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