Classical Mechanics (Kibble & Berkshire) · Problem 4.23

Linear oscillations — T.W.B. Kibble, F.H. Berkshire

Physics — Mechanics · projectile-parabola-of-safety

Problem

A particle is projected from the origin with speed uu in a direction making an angle α\alpha with the horizontal. The motion takes place in the (x,z)(x,z)-plane, where OzOz points vertically upwards. If the projection speed uu is fixed, show that the particle can be made to pass through the point (a,b)(a,b) for some choice of α\alpha if (a,b)(a,b) lies below the parabola z=u22g(1−g2x2u4).z = \frac{u^2}{2g}\left(1 - \frac{g^2x^2}{u^4}\right). This is called the parabola of safety. Points above the parabola are `safe' from the projectile. An artillery shell explodes on the ground throwing shrapnel in all directions with speeds of up to 30 m s−130\text{ m s}^{-1}. A man is standing at an open window 20 m above the ground in a building 60 m from the blast. Is he safe? [Take g=10 m s−2g = 10\text{ m s}^{-2}.]

Browse the problem library