Help with Mechanics questionsWatch
Q. Alan throws a stone vertically upwards.
i) He throws it at 25ms^-1. Show that it takes 5.1 seconds to return to his hand. What modelling assumptions have you made for this calculation?
Alan throws the stone, again vertically upwards, at ums^-1. Answers for the rest of this question should be given in terms of u and g.
ii) Find an expression for h, the maximum height above the point of projection reached by the stone.
iii) Find an expression for the height of the stone above the point of projection t seconds after reaching the highest point.
iv) When the stone is at its maximum height, Alan throws another stone vertically upwards, also at ums^-1.
Prove that the stones cross in the air three-quarters of the way up to the highest point above the point of projection.
for modelling assumptions could say not including air resistance, or wind, modelling stone as a particle
for ii) maximum height is when v=0 so v^2=u^2+2as so 0=u^2 -2gs so s=(u^2)/2g s= max h
iii) use s=ut+0.5at^2 again where s=h t=t, u=u, and a=-g
iii) you need to use simultaneous equations so
STONE 1: s=? u=u v=n/a a=-g t=t
STONE 2: s=? u=u v=n/a a=-g t= t+ (25/0.5g)/2 so t= t+25/g
then s=ut+0.5at^2 and equate
so ut-0.5gt^2=u(t+25/g)-0.5g(t+25/g)^2 solve for t and then use t to find the s where they meet which should be 0.75x (u^2)/2g