projectile motion (1 Viewer)

[redd]

help please

A vertical building stands with base O on horizontal ground. A and B are two points on the building vertically above each other such that A is 4h metres above O and B is h metres above O. A particle is projected horizontally with speed U ms from A and 10 seconds later a second particle is projected horizontally with speed V ms from B. the two particles hit the ground at the same point and at the same time.

derive the expressions for the horizontal and the vertical displacements relative to O of each particle t seconds after the first particle is projected

find the time of flight of each particle

show that the ratio of the velocities are in 2:1 ratio

 

[rukie]

Dude thats weird, i keep getting the same value for time but its buggering the ratio.

For A
y = -g/2*t^2 + 4h
x = ut

For B
y = -g/2*t^2 + 10gt + h
x = v*(t-10)

Time = 3h/10g

so yeah, when i equate the horzontal displacements to get the ratio i got

u/v = (3h -100g)/3h

So in any case its wrong so ignore it, just thought id give it a shot

 

[redd]

ive managed to get ii and iii

for ii just let

y2 = 0

y1= 0

then eliminate h

for the1st part I cheated by looking at answer

dunno how they got the expressions

 

[KFunk]

For A
y = 4h - 1/2.gt² (1)

For B
y = h - 1/2.g(t - 10)² (2)


If we let y=0 when t = t₀ then

from (1) ---> t₀ = √(8h/g) = 2√(2h/g) ....... (3)
from (2) ---> t₀ = 10 + √(2h/g) ......................(4)
equating (3) with (4) we get 2h/g = 100 ---> h/g = 50 (5)

subbing (5) back into (3) we get t₀ = 20

For A (x = Ut) and for B (x = V(t - 10)). These are equal when t = t₀ = 20

Ut₀ = V(t₀ - 10)

20U = 10V , ∴ V/U = 2 , i.e. the ratio is 2:1


There's probably a quicker way to go about it but this seems to work out.

 

[tim_dawborn]

solutions

yea that is the best way that i can see to do it . attached is a nicer copy of the solutions in case you are interested

 

[redd]

thats really nice sol but just a query about the derivation for particle B

since its projected 10 seconds after wouldn't v = V -10 or something (or v = 10+V ?)

 

[tim_dawborn]

thats really nice sol but just a query about the derivation for particle B

since its projected 10 seconds after wouldn't v = V -10 or something (or v = 10+V ?)

There is a mistake in my solutions --> particle B x displacement equation is meant to be <b>x = V(t - 10)</b>

you see how the time for particle B is (t - 10), that accounts for the ten second gap of teh particles.

if it was v = V - 10, that is constantly shaving 10 ms^-1 from the velocity, which is not what it is saying, it is saying 10 seconds

 

[redd]

ah yes, i see i see

thanks alot dude