Hi,

I am having some trouble with the following Qs.

I prefer to tackle these type of projectile Qs using the vector i,j notation.

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A stone is projected from a point O on a cliff with a speed of 20 ms^-1 at an angle of elevation of 30 degrees. T seconds later the angle of depression of the stone from O is 45 degrees. Find the value of T.

My workings for Q1

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Let r ' ' = -9.8 (using r ' ' to show r double dot, r ' to show r dot)

r ' = -9.8t + A i +B j

but, r ' = 20cos30i + 20sin30j when t=0

so, A=20cos30, B=20sin30

hence, r ' = (20cos30)i + (20sin30 -9.8t)j

r = (20tcos30)i + (20t sin30 -4.9t²)j

not quite sure of what to do next, got muddled up... something along the line of if 20cos45i + 20sin45j etc etc... but I need some help.

Q2

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A particle is projected from a point on level ground. Its maximum height is 20m and hits the ground 200m from its point of projection. Find the time of flights and the angle of projection.

My workings for Q2

============

I used similar notation as in Q1.

I got into lots of variables in this Q but was not able to eliminate them and made a big mess of it. Hope someone can help.

Thanks in advance.

A very tired person.... (look at the time of the post!!)

HAL2001

Your method of using vector notation is the same as considering the motion in components, except you are considering both components simultaneously. It might be clearer to consider them separately, and to use the constant acceleration equations, for a few problems and then go back to treating them simultaneously once you have more of a feel for what is going on. You may not like to use the constant acceleration formulae, but you will only be rederiving them each time you do a problem.

As for Q2, then.

Horizontally:

Neglecting air resistance etc. there are no forces acting on the particle,

so using

s = u t +

1 2

a t2

we have x = u t cosq+ 0

(where x is horizontal displacement, q is angle of projection above horizontal)

Þ 200 = u T cosq

(where T is time of flight)

Vertically:

There is only a gravitational force acting on the particle, so using the same equation as before we have

y = u t sinq-

1 2

g t2

(where y is vertical displacement, taking upwards direction as positive)

Þ20 =

1 2

u T sinq- (1/8)g T2

,

using the symmetry of the motion i.e. the particle takes the same time to get up to its maximum height as it does to get back down ( N.B. this would not apply if we accounted for air resistance).

Also

0 = u T sinq-

1 2

g T2

(the vertical displacement is zero when the particle lands)

Thanks Barkley.

Q2, I tried the questions using i,j notation, it ended up with the same equations as you have. As so many variables started popping up and it was quite late as well, I got a little confused. The trick was to obtain 3 equations for the 3 variables, u, T and q. I don't know why, but I guess using i, j notation is just got a liking to myself, I still use the constant acceleration formulas for when particles are not projected at an angle.

Does anyone find that when using i,j notation less thinking is required? The answers just seem to pop out easier without much thought. I feel that the constant acceleration formulas require a tad more thinking from the user. Just my two cents.

Q1, the notation that I used r.. etc was slightly confusing when reading off a screen... but I think I get the general drift of the projective v/h - consider them separately i guess.

Thanks once again for you help.