Question

james can u come here??

Answer 1

Here

Answer 2

yey

Answer 3

wl today i hav some physics qusestions

Answer 4

ok

Answer 5

start?

Answer 6

Sure ... I'm making coffee now so I'll be good this morning, my morning that is!

Answer 7

Go ahead. Write the problem here or in a new question.

Answer 8

A ball is projected so as to clear 2 walls,the first of height a at a distance b from the point of projection and the second of height b and at a distance a from point of projection.Show that the range on the horizontal plane is \[a ^{2}+a timesb+b ^{2}\div a+b\]

Answer 9

a ^{2}+a timesb+b ^{2}\div (a+b)

Answer 10

I.e.,
\[\frac{a^2 +ab + b^2}{a+b}\]
What does the question mean by range? How far the ball travels?

Answer 11

ya horizontal distance covered by projectile

Answer 12

Let x be the horizontal axis and y the vertical height.

Then the ball traces a parabola, y = y(x),

\[y(x) = px^2 + qx + r\]
for some constants p, q and r

Let's assume we launch the projectile from the origin. Then y(0) = 0. Hence r = 0. Hence y(x) simplifies to
\[ y(x) = px^2 + qx \]

Now let us assume the ball just clears both walls.

Then
\[y(a) = b \ \text{ and } \ y(b) = a\]
That is,
\[b = a^2.p + a.q \ \text{ and } \ a = b^2.p + b.q \]

This is a system of two simultaneous equations in p and q. Solve them for p and q in terms of a and b.

Then you will have an expression for y(x) only in terms of x, and a and b.

Answer 13

the range is the projection onto the horizontal x-axis

Answer 14

The distance covered by the projectile will be the non-trivial solution of y(x) = 0.

Answer 15

oh i am only high school i cant use the equation of parabola

Answer 16

is thr othr method?

Answer 17

Yes you can.

Answer 18

plz tel

Answer 19

learn now

Answer 20

Haven't you discussed how in general ideal projectiles trace out parabolas?

Answer 21

If not, let just prove it very quickly. It's easy.

Answer 22

oh i needs to use equation of projectile and solve

Answer 23

Suppose a projectile is launch with speed v0 at an angle theta to the horizontal

Then resolving in x, y directions, the v0 in the x direction v0x = v cos theta and in the y-direction, v0y = v sin theta

Then vy(t) = v0y - gt
vx(t) = v0x

....

Answer 24

What equations do you have then?