need help! projectile motion (1 Viewer)

[j@son]

1) A particle is projected to just clear two walls of height 7 metres and distant 7 metres and 14 metres from the point of projection. Prove that if A is the angle of projection then tanA=3/2. Prove that if the walls are h metres high and distant b metres and c metres from the point of projection then
tanA=h(b-c)/bc.

2) A particle projected from a point meets the horizontal plane through the point of projection after travelling a horizontal distance a, and in the course of its projectory attains a greatest height b above the point of projection. Find the horizontal and vertical components of the velocity in terms of a and b. Show that when it has described a horizontal distance x, it has attained a height of 4bx(a-x)/a^2.

3) A football, kicked at 16m/s, just passes over a cross-bar 4 metres high and 16 metres away. Show that if A is the angle of projection,
5tan^2A-16tanA+9=0.

thanks.

 

[insert-username]

The working out for all these are very long, so I won't be doing them in full.

1) A particle is projected to just clear two walls of height 7 metres and distant 7 metres and 14 metres from the point of projection. Prove that if A is the angle of projection then tanA=3/2. Prove that if the walls are h metres high and distant b metres and c metres from the point of projection then
tanA=h(b-c)/bc.

Firstly, the cartensian equation for the path of the projectile is: y = xtan@ - gx²(1+tan²@)/2V²

When x is 7, y = 7

Thus 7 = 7tan@ - 245/v²(1 - tan²@)

And v² = [245(1 - tan²@)]/7(tan@ - 1)


When x is 14, y = 7

So 7 = 14tan@ - 980/²v(1 - tan²@)

Now substitute in for v², and solve for tan@, which will equal 3/2.

You do the second half of the question in a different way, which I can't remember. I shall post it if I remember it.

3) A football, kicked at 16m/s, just passes over a cross-bar 4 metres high and 16 metres away. Show that if A is the angle of projection,
5tan^2A-16tanA+9=0.

V = 16

And when x = 16, y = 4

So 4 = 16tanA - (10.256/2.256)(1 + tan²A)

4 = 16tanA - 5(1 + tan²A)

4 = 16tanA - 5 - 5tan²A

Therefore 5tan²A - 16tanA + 9 = 0


I_F