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Report Thread starter 3 years ago
Please can you help me do these questions, the teacher said to leave the questions as he believed them to be too complex I think for our level atm, but I want to get a step ahead.

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.
Badges: 11
Report 3 years ago
for the first part just use suvat equations in the vertical plane, so taking upwards to be positive direction you would have s=0, u=25, v not applicable, a=-9.8 t=t and then could use s=ut+0.5at^2 to get 0=25t-4.9t^2 so t=25/4.9 =5.1
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

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