Asked by Anonymous
A 0.19 kg ball of dough is thrown straight up into the air with an initial speed of 13 m/s.The acceleration of gravity is 9.81 m/s.
b) What is its momentum halfway to its maximum height on the way up?
Answer in units of kg ยท m/s.
b) What is its momentum halfway to its maximum height on the way up?
Answer in units of kg ยท m/s.
Answers
Answered by
bobpursley
max height:(velocity is zero)
vf^2=vi^2+2ad
0=13^2-2g
max height=13^2/2g
halfway=13^2/4g
vf^2=vi^2+2ad'
vf^2=13^2 -2g*13^2/4g
v half way= sqrt(13^2/2)
momentum half way=mass times vhalfway.
vf^2=vi^2+2ad
0=13^2-2g
max height=13^2/2g
halfway=13^2/4g
vf^2=vi^2+2ad'
vf^2=13^2 -2g*13^2/4g
v half way= sqrt(13^2/2)
momentum half way=mass times vhalfway.