projectile motion question (1 Viewer)

[Doornub]

hey guys,

just need help with this question

An object falling in a vertical line passes a window 3 m high in 1/6 second. Find the distance above the top of the window from which the object was let fall.

Plz use the projectile motion maths formulas..

thanks in advance

 

[gurmies]

y: = -g

y. = -gt

y = -gt^2/2 + C

When t = 1/6, y = 3

C = 3 + 5(1/6)^2

= 113/36 m

113/36-3 = 5/36 metres above the window?

 

[Doornub]

the answer is actually 15.06m =/

 

[gurmies]

the answer is actually 15.06m =/

Yeah, I'm not sure - probably did something wrong.

 

[jet]

Well, you're missing a constant of integration from the velocity.

 

[oly1991]

the only way to this question is the way gurmies did it

i cannot see another way

i may be wrong though....

 

[kaz1]

hey guys,

just need help with this question

An object falling in a vertical line passes a window 3 m high in 1/6 second. Find the distance above the top of the window from which the object was let fall.

Plz use the projectile motion maths formulas..

thanks in advance

I remember this in the old Jones/Couchman book. You're supposed to use the Physics method (it says so in the book).

s=ut+ 0.5at²
3=u/6 + (4.9)(1/36)
u=17.18

now let u be final velocity(v)

v²=u² +2as
17.18²=0+19.6s
s=15.06m

I'm pretty sure you were meant to use the physics method in this exercise.

 

[gurmies]

Well, you're missing a constant of integration from the velocity.

I assumed when t = 0, v =0 - seems silly in retrospect.