Projectile Motion - A Question (1 Viewer)

[Shadowdude]

I tried my "professional" way of solving a question to the effect of:

A cannon is placed on a hill 50m above the sea, at an angle of 45 to the horizontal. If a cannonball is fired at 25m/s at the angle, what is the full range of the flight?
​

Now what I did was I calculated normal range "d = u(x) . t". Then, I figured that if the cannonball was launched at 45, as long as it landed on a parallel surface to the original place of projection - it too would land at an angle of 45.

I then applied 'tan' (the trigonometric ratio), and got an answer 0.4m from the actual answer.

How do I do it 'properly' and get rid of that 0.4m error?

Sorry that I couldn't find the actual question...

 

[Pwnage101]

A cannon is placed on a hill 50m above the sea, at an angle of 45 to the horizontal. If a cannonball is fired at 25m/s at the angle, what is the full range of the flight?

Assume that where the cannon ball lands is 50 m below the top of the hill.

U=25
Ux=25cos45
Uy=25sin45

Method 1: Long way

Thus Ux=Uy=17.6776695..... [This is only the case since the angle is 45 degrees]
Let T1 be the time taken for the ball to move from its launch position to maximum height, and T2 be the time taken for the ball to move from maximum height to its landing position.

We require T*=T1+T2

Now
we know Vy=0 at max height

thus using Vy=Uy+AyT1

0=25sin45+(-9.8)t

thus T1=(-25sin45)/(-9.8)=1.80384...

In this time the ball has travelled vertically by:
Sy=UyT1+0.5Ay(T1)^2
=25sin45(1.80384)-4.9(1.80384)^2
=15.9438...m

Thus now the ball is 15.9438+50=65.9438 m above ground

now for T2

Sy=UyT2+0.5Ay(T2)^2

-65.9438=0T2-4.9(T2)^2

(T2)^2=(-65.9438/-4.9)
T2=(-65.9438/-4.9)^(1/2)****

****^(1/2) is equivalent to 'square root'

T2=3.6685...

Thus Total time of flight =T*=T1+T2=1.80384+3.6685=5.4723...s

Range=Sx=UxT*=25cos45*5.4723=96.738386...=96.7m

Method 2: SHort way, required to solve a quadratic equation using the quadratic formula

Sy=Uyt+0.5Ay(t)^2

Now we can take Sy to be -50, and the balla chieves this in T* seconds thus:

-50=(25sin45)T*-4.9(T*)^2

Rearrange:

4.9(T*)^2-25sin45T*-50=0

This is a quadratic equation in T*, THUS using quadtratic formula:

T*=[-b+(b^2-4ac)^(1/2)]/2a---> only take positive square root as the other quare root is negative and we are not intersted in this

T*=[-(-25sin45)+((25in45)^2-4*4.9*-50)^(1/2)]/(9.8)
=5.4723...

and proceed as above

 

[Shadowdude]

Oh so that's how you do it! Thanks for clearing that up.

So I had to use more correct values for y-displacement... I see.

 

[lordinance]

A cannon is placed on a hill 50m above the sea, at an angle of 45 to the horizontal. If a cannonball is fired at 25m/s at the angle, what is the full range of the flight?

Assume that where the cannon ball lands is 50 m below the top of the hill.

U=25
Ux=25cos45
Uy=25sin45

Method 1: Long way

Thus Ux=Uy=17.6776695..... [This is only the case since the angle is 45 degrees]
Let T1 be the time taken for the ball to move from its launch position to maximum height, and T2 be the time taken for the ball to move from maximum height to its landing position.

We require T*=T1+T2

Now
we know Vy=0 at max height

thus using Vy=Uy+AyT1

0=25sin45+(-9.8)t

thus T1=(-25sin45)/(-9.8)=1.80384...

In this time the ball has travelled vertically by:
Sy=UyT1+0.5Ay(T1)^2
=25sin45(1.80384)-4.9(1.80384)^2
=15.9438...m

Thus now the ball is 15.9438+50=65.9438 m above ground

now for T2

Sy=UyT2+0.5Ay(T2)^2

-65.9438=0T2-4.9(T2)^2

(T2)^2=(-65.9438/-4.9)
T2=(-65.9438/-4.9)^(1/2)****

****^(1/2) is equivalent to 'square root'

T2=3.6685...

Thus Total time of flight =T*=T1+T2=1.80384+3.6685=5.4723...s

Range=Sx=UxT*=25cos45*5.4723=96.738386...=96.7m

Method 2: SHort way, required to solve a quadratic equation using the quadratic formula

Sy=Uyt+0.5Ay(t)^2

Now we can take Sy to be -50, and the balla chieves this in T* seconds thus:

-50=(25sin45)T*-4.9(T*)^2

Rearrange:

4.9(T*)^2-25sin45T*-50=0

This is a quadratic equation in T*, THUS using quadtratic formula:

T*=[-b+(b^2-4ac)^(1/2)]/2a---> only take positive square root as the other quare root is negative and we are not intersted in this

T*=[-(-25sin45)+((25in45)^2-4*4.9*-50)^(1/2)]/(9.8)
=5.4723...

and proceed as above

Why S=-65.9438m?

 

[adomad]

Why S=-65.9438m?

delta y is the change in vertical distance. at apeggee, where the vertical velocity is zero and is at the max of its motion, the projectile will be 15.9348m higher than the top of the cliff ( 65.9348m) in terms of vertical displacement. now when we are calculating the time for the projectile to drop from this new height to the floor, we must get delta y to use the equation delta y = ut + 0.5*a*t^2 or as some people know it, S= ut+1/2 a t^2

the change in vertical distance = final distance - initial distance. hence 0 (because we are at the base of the cliff) - 65.9348 = -65.9348m

 

[Pwnage101]

delta y is the change in vertical distance. at apeggee, where the vertical velocity is zero and is at the max of its motion, the projectile will be 15.9348m higher than the top of the cliff ( 65.9348m) in terms of vertical displacement. now when we are calculating the time for the projectile to drop from this new height to the floor, we must get delta y to use the equation delta y = ut + 0.5*a*t^2 or as some people know it, S= ut+1/2 a t^2

the change in vertical distance = final distance - initial distance. hence 0 (because we are at the base of the cliff) - 65.9348 = -65.9348m

a shorter way to solve the problem is to do this

delta y = ut + 0.5 at^2

Delta y = -50 m because it was fired at a height of 50 m and then went down to the water which is 0m above the sea level.

u= 25sin45 = 17.68ms-1

plug them in and you get a quadratic 4.9t^2 -17.68-50=0, use you calculator to solve it.

you will get one positive answer and one negative answer, and if you have solved it correctly the positive answer will be bigger that the negative one .

hence t = 5.476 or -1.86 .'. t= 5.476

then range= ut =5.476*17.68 = 96.763m

Yes, i provided that (shorter) method above, in my original response ^^^. It just requires use of the quadratic formula.

RE: why Sy=-65.9438, it is because i am finding T2, which i defined clearly to be the "the time taken for the ball to move from maximum height to its landing position" - its maximum height was calculated to be 15.9438...m above its initial position, which itself is 50 m above its final (vertical) position, thus displacement =-65.9438 [negative because it is going down , and convention is tot ake up as positive.

 

[adomad]

Yes, i provided that (shorter) method above ^^^. It just requires use of the quadratic formula.

opps, didn't see that... will remove

 

[xXmuffin0manXx]

mm i checked it it..

but..really..

your final two lines should resemble something like this..

delta(x) = 96.73830683m
= 97m (2 sig figs)

 

[Pwnage101]

mm i checked it it..

but..really..

your final two lines should resemble something like this..

delta(x) = 96.73830683m
= 97m (2 sig figs)

True, if indeed the question used only 2 significant firgures, the final answer should be given to two significant figures.

I did it to 1 decimal place because the OP stated specifically they got an 0.4m error:

I then applied 'tan' (the trigonometric ratio), and got an answer 0.4m from the actual answer.

How do I do it 'properly' and get rid of that 0.4m error?

.

 

[xXmuffin0manXx]

True, if indeed the question used only 2 significant firgures, the final answer should be given to two significant figures.

I did it to 1 decimal place because the OP stated specifically they got an 0.4m error:


.

oh right...yeah i didnt actualyl read your post..i just read adomads..and he knows how stingy i am with sig figs

 

[adomad]

oh right...yeah i didnt actualyl read your post..i just read adomads..and he knows how stingy i am with sig figs

true lol

 

[Pwnage101]

oh right...yeah i didnt actualyl read your post..i just read adomads..and he knows how stingy i am with sig figs

Good to be like that - many students don't understand/use sig figs, so you've got a definite advantage.

 

[GINN]

I tried my "professional" way of solving a question to the effect of:

A cannon is placed on a hill 50m above the sea, at an angle of 45 to the horizontal. If a cannonball is fired at 25m/s at the angle, what is the full range of the flight?
​

Now what I did was I calculated normal range "d = u(x) . t". Then, I figured that if the cannonball was launched at 45, as long as it landed on a parallel surface to the original place of projection - it too would land at an angle of 45.

I then applied 'tan' (the trigonometric ratio), and got an answer 0.4m from the actual answer.

How do I do it 'properly' and get rid of that 0.4m error?

Sorry that I couldn't find the actual question...

oh very typical of elitist pricks who want to use a 'professional' approach.

a realistic approach involves taking differential models and taking into account drag force.