Question

pleas ehelp:)

Answer 1

http://screencast.com/t/8Ty4aIuTIg

Answer 2

?

Answer 3

@Vincent-Lyon.Fr

Answer 4

Unable to download the link.

Answer 5

http://screencast.com/t/8Ty4aIuTIg

Answer 6

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w03_qp_5.pdf

5th question
ii) iii)

Answer 7

@Vincent-Lyon.Fr @satellite73

Answer 8

\[\frac{ dv }{ 5-v }=\frac{ dt }{ 10 }\]
\[\int\frac{ dv }{ 5-v }=\int\frac{ dt }{ 10 }\]
\[-\ln (5-v)=\frac{ t }{ 10 }\]
\[\huge 5-v=e^{-\frac{ t }{ 10 }} \implies v=5-e^{-\frac{ t }{ 10 }}\]

Answer 9

http://screencast.com/t/8Ty4aIuTIg

This is the question bro!

Answer 10

\[v_y^2=v_0^2+2a y \implies y=\frac{ v_y^2 }{ 2a }=\frac{ (v \sin \alpha)^2 }{ 2g }=\frac{ 20^2 \sin^2 \alpha }{ 2*10 }=7.2\]

Answer 11

\[\alpha=36,9\]

Answer 12

\[\alpha=36.9\]

Answer 13

the second part is the problem

Answer 14

first we find the time
\[t=?\]
\[v_{fy}=v_{iy}+at\]
\[0=v_{iy}-gt\]
\[20 \sin 36.9=10 t\implies t=1.2\]

speed on hitting the wall is \[v_x=20 \cos \alpha=20 \cos 36.9=20*0.8=16 \]
half \[v_x=8\]

\[\huge x=v_xt=1.2*16=9.6m\]

Answer 15

oh ok thanks:)

Answer 16

\[\tan \theta=\frac{ v_y }{ v x }=\frac{ 10*1.2 }{ 10 }=.1.2 \implies \theta=56\]