Air resistance kv² - projected vertically upward

5 A particle \(P\) of mass 0.5 kg is projected vertically upwards from a point on a horizontal surface. A resisting force of magnitude \(0.02 v ^ { 2 } \mathrm {~N}\) acts on \(P\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the upward velocity of \(P\) when it is a height of \(x \mathrm {~m}\) above the surface. The initial speed of \(P\) is \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that, while \(P\) is moving upwards, \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 10 - 0.04 v ^ { 2 }\).
2. Find the greatest height of \(P\) above the surface.
3. Find the speed of \(P\) immediately before it strikes the surface after descending.

5 A particle \(P\) of mass 0.5 kg is projected vertically upwards from a point on a horizontal surface. A resisting force of magnitude \(0.02 v ^ { 2 } \mathrm {~N}\) acts on \(P\), where \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) is the upward velocity of \(P\) when it is a height of \(x \mathrm {~m}\) above the surface. The initial speed of \(P\) is \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Show that, while \(P\) is moving upwards, \(v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 10 - 0.04 v ^ { 2 }\).
2. Find the greatest height of \(P\) above the surface.
3. Find the speed of \(P\) immediately before it strikes the surface after descending.

5 A particle \(P\) of mass \(m \mathrm {~kg}\) is projected vertically upwards through a liquid. Student \(A\) measures \(P\) 's initial speed as \(( 8.5 \pm 0.25 ) \mathrm { m } \mathrm { s } ^ { - 1 }\) and they also record the time for \(P\) to attain its greatest height above the initial point of projection as over 3 seconds.
In an attempt to model the motion of \(P\) student \(B\) determines that \(t\) seconds after projection the only forces acting on \(P\) are its weight and the resistance from the liquid. Student \(B\) models the resistance from the liquid to be of magnitude \(m v ^ { 2 } - 6 m v\), where \(v\) is the speed of the particle.

1. (a) Show that \(\frac { \mathrm { d } t } { \mathrm {~d} v } = - \frac { 1 } { ( v - 3 ) ^ { 2 } + 0.8 }\).
   (b) Determine whether student \(B\) 's model is consistent with the time recorded by \(A\) for \(P\) to attain its greatest height. After attaining its greatest height \(P\) now falls through the liquid. Student \(C\) claims that the time taken for \(P\) to achieve a speed of \(1 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) when falling through the liquid is given by $$- \int _ { 0 } ^ { 1 } \frac { 1 } { ( v - 3 ) ^ { 2 } + 0.8 } \mathrm {~d} v$$
2. Explain why student \(C\) 's claim is incorrect and write down the integral which would give the correct time for \(P\) to achieve a speed of \(1 \mathrm {~ms} ^ { - 1 }\) when falling through the liquid.

2. At time \(t = 0\), a particle \(P\) of mass \(m\) is projected vertically upwards with speed \(\sqrt { \frac { g } { k } }\), where \(k\) is a constant. At time \(t\) the speed of \(P\) is \(v\). The particle \(P\) moves against air resistance whose magnitude is modelled as being \(m k v ^ { 2 }\) when the speed of \(P\) is \(v\). Find, in terms of \(k\), the distance travelled by \(P\) until its speed first becomes half of its initial speed.
(9)

2. At time \(t = 0\), a particle \(P\) of mass \(m\) is projected vertically upwards with speed \(\sqrt { \frac { g } { k } }\), where \(k\) is a constant. At time \(t\) the speed of \(P\) is \(v\). The particle \(P\) moves against air resistance whose magnitude is modelled as being \(m k v ^ { 2 }\) when the speed of \(P\) is \(v\). Find, in terms of \(k\), the distance travelled by \(P\) until its speed first becomes half of its initial speed.

12 A bullet of mass 0.0025 kg is fired vertically upwards from a point \(O\). At time \(t \mathrm {~s}\) after projection the speed of the bullet is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the resistance to motion has magnitude \(0.00001 v ^ { 2 } \mathrm {~N}\).

1. Show that, while the bullet is rising, $$250 \frac { \mathrm {~d} v } { \mathrm {~d} t } = - 2500 - v ^ { 2 }$$
2. It is given that the speed of projection is \(350 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find
   1. the time taken after projection for the bullet to reach its greatest height above \(O\),
   2. the greatest height above \(O\) reached by the bullet.

10 A small body of mass \(m\) is thrown vertically upwards with initial velocity \(u\). Resistance to motion is \(k v ^ { 2 }\) per unit mass, where the velocity is \(v\) and \(k\) is a positive constant. Find, in terms of \(u , g\) and \(k\),

1. the time taken to reach the greatest height,
2. the greatest height to which the body will rise.

12 A particle of mass 0.6 kg is projected vertically upwards from horizontal ground, with initial speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The upwards velocity at any instant is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), and the displacement is \(x \mathrm {~m}\). Air resistance is modelled by a force \(0.024 v ^ { 2 } \mathrm {~N}\) acting downwards.

1. Show that \(v\) and \(x\) satisfy the differential equation $$v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 10 - 0.04 v ^ { 2 } .$$
2. Find the value of \(u\) if the maximum height reached is 50 m .

A ball of mass \(m\) is thrown vertically upwards from the ground. When its speed is \(v\) the magnitude of the air resistance is modelled as being \(mkv^2\), where \(k\) is a positive constant. The ball is projected with speed \(\sqrt{\frac{g}{k}}\). By modelling the ball as a particle,

1. find the greatest height reached by the ball. [9]
2. State one physical factor which is ignored in this model. [1]