How to do this projectile question (1 Viewer)

[kooltrainer]

24) Find the speed and direction of a particle which, when projected from a point 15 m above the horizontal ground, just clears the top wall 26.25 m high and 30 m away.

 

[conics2008]

ok use your catersion formula to find V and theta... make sure you derive them dont just use them.. take g=-9.8

.. if you still cant.. i will post up answer.. thank you

 

[kooltrainer]

tried it... didnt work out...........

 

[conics2008]

ok i post now.

im struggling to get v^2 out of the way ??

 

[kooltrainer]

yeh.. its pretty hard..
answer says velocity should be 25 m/s and angle shuld be about 37 degrees...

 

[ronnknee]

ok use your catersion formula to find V and theta... make sure you derive them dont just use them.. take g=-9.8

.. if you still cant.. i will post up answer.. thank you

In Maths, you should take g = 10, otherwise stated

 

[Affinity]

From the question we know that at some particular time T , y'(T) = 0, y(T) = 26.5 and x(T) = 30

1. show that [(y')^2 / 2] + gy is constant
2. from that we can deduce that y'(0) = sqrt(2*11.25*g) = sqrt(22.5g)
3. ofcourse y'(t) = y'(0) - gt, so in particular, 0=y'(T) = y'(0) - gT
so T = y'(0)/g = sqrt(22.5/g)
4. x(T) = x'(0)*T so x'(0) = 30/sqrt(22.5/g) = sqrt(900g/22.5)

5. so initial speed is about 24.6 (taking g = 9.8), initial angle is about 36.86 degrees

fixed the error sorry for the confusion :S

 

[Affinity]

In Maths, you should take g = 10, otherwise stated

In maths if you aren't given g then you write g as g.

 

[minijumbuk]

I tried this question, but I didn't get the same answer as Affinity. Can someone please tell me where I went wrong?

Firstly, I minused the distance between top of the wall and where the projectile is shot. So, I got: Range = 30m, max height = 11.25m

Then, I did this:
a_x=0
v_x= V cosѲ
x= Vt cosѲ +C
When t=0, x=0, therefore C=0
x= Vt cosѲ
_____________________________
a_y= -9.8
v_y= -9.8t +C
t=0, v_y= VsinѲ
Therefore v_y = -9.8t +VsinѲ
y= -4.9t²+VtsinѲ + C
t=0, y=0, C=0
y= -4.9t²+VtsinѲ
________________________________

When v_y=0, t=max
-9.8t + VsinѲ = 0
t= VsinѲ/9.8
When t =VsinѲ/9.8 , y=11.25
11.25 = -4.9(VsinѲ/9.8)² + V²sin²Ѳ/9.8
= -5V²sin²Ѳ/98 + 10V²sin²Ѳ/98
= (5V²sin²Ѳ ) /98
V²sin²Ѳ = 220.5 --------1

When t= VsinѲ /9.8, x=30
30 = (V²sinѲ cosѲ ) /9.8
294 = V²sinѲcosѲ
294sinѲ = V²sin²ѲcosѲ
From 1,
294tanѲ = 220.5
tanѲ = 3/4
Ѳ=36 degrees and 52 minutes

V²sin²Ѳ = 220.5
V² = 220.5 / sin²(36degrees52mins)
V = 24.75 m/s

 

[minijumbuk]

Wow, I got the same as 3unitz =\

How do I do it the cartesian way? I never learnt of such a method =\
So I have to derive stuff every single time xD

 

[Affinity]

oh I made a mistake there.. 26.25 - 15 is 11.25 not 11.5 my bad..
Anyway, the main purpose of that post was to show that one can use the familiar idea of energy from physics.

 

[independantz]

I tried this question, but I didn't get the same answer as Affinity. Can someone please tell me where I went wrong?

Firstly, I minused the distance between top of the wall and where the projectile is shot. So, I got: Range = 30m, max height = 11.25m

Then, I did this:
a_x=0
v_x= V cosѲ
x= Vt cosѲ +C
When t=0, x=0, therefore C=0
x= Vt cosѲ
_____________________________
a_y= -9.8
v_y= -9.8t +C
t=0, v_y= VsinѲ
Therefore v_y = -9.8t +VsinѲ
y= -4.9t²+VtsinѲ + C
t=0, y=0, C=0
y= -4.9t²+VtsinѲ
________________________________

When v_y=0, t=max
-9.8t + VsinѲ = 0
t= VsinѲ/9.8
When t =VsinѲ/9.8 , y=11.25
11.25 = -4.9(VsinѲ/9.8)² + V²sin²Ѳ/9.8
= -5V²sin²Ѳ/98 + 10V²sin²Ѳ/98
= (5V²sin²Ѳ ) /98
V²sin²Ѳ = 220.5 --------1

When t= VsinѲ /9.8, x=30
30 = (V²sinѲ cosѲ ) /9.8
294 = V²sinѲcosѲ
294sinѲ = V²sin²ѲcosѲ
From 1,
294tanѲ = 220.5
tanѲ = 3/4
Ѳ=36 degrees and 52 minutes

V²sin²Ѳ = 220.5
V² = 220.5 / sin²(36degrees52mins)
V = 24.75 m/s

I don't know if i'm missing something here, but shouldn't the second be 2x the first one ( in bold) as the max height occurs at half the time of flight, thus when the range occurs you need to multiply it by 2?

 

[minijumbuk]

No, because I found when vertical velocity is 0. It doesn't matter if it's half time or full time. As long as it's at v_y=0, it is maximum.