So i got this question and i managed to do part i and ii however i'm struggling to do part iii, can someone show me how to do this.
A particle P is projected vertically upwards, from horizontal ground, with speed 8.4 m s−1
(i) Show that the greatest height above the ground reached by P is 3.6 m. [3]A particle Q is projected vertically upwards, from a point 2 m above the ground, with speed um s−1
Solution:
The greatest height above the ground reached by Q is also 3.6 m.(ii) Find the value of u. [2]It is given that P and Q are projected simultaneously.(iii) Show that, at the instant when P and Q are at the same height, the particles have the same speedand are moving in opposite directions. [6]
Take the ground level as being the zero for displacement and measure displacements as positive upwards.
(i) The equation of motion for P is: \( a = s'' = -g \) m/s^2.
Integrating once we get: \( v = s' = -g t + 8.4 \) m/s
and again:\( s = - (g t^2)/2 + 8.4 t \)m
The maximum height is achieved when \(v=0\), so occurs at \(t_{mx}= 8.4/g \) seconds, substituting this into the displacement equation we get the maximum displacement (maximum height) is:
\( \phantom{11} s_{mx} = - (g (8.4)^2)/2 + 8.4 (8.4/g) = (8.4)^2 / (2g) \approx 3.59 \approx 3.6 \) m,
(ii) For Q we have:\( a=-g,\ v = -gt +u\) and \(s= -(g t^2)/2 + ut + 2 \).
Q reaches its maximum height when \(v = 0\), so at \(t_{mx} = u/g\) s. At this time the displacement is:
\( \phantom{11} s_{mx} = - g (u/g)^2 /2 +u^2/g +2 = u^2/(2g) +2 = 3.6\) m
So now we can solve for \(u\).
(iii) You now have the displacement equations for both P and Q and the two particles are at the same height at the time when these two displacements are equal. So equate the displacements which will give you a quadratic in t. You want the positive root of this which gives the time when they are at the same height. Now plug this time into the two velocity equations ...
