Question

a rock fliesupward with velocity 26 m/s(52 degree above horizontal).rock rises to maximum height and lands 12m above initial position.(acceleration due to gravity on this planet is 3.7 m/s squared
calculate a)max.height b)time rock is in flight c)range of rock

Answer 1

you had a question about this one?

Answer 2

yea ..m not getting the max. height right...its 57m....m getting 61m..

Answer 3

wat i did was vfy at max height is 0
vfy=viy-at
0=26 sin 52-3.4t
t=6.03s
then
h=viyt-1/2g[t ^{2}\]
\[h=26 \sin 52(6.03)-\frac{ 3.4 }{ 2 }(6.03)^{2}\]
h=61.73 m

Answer 4

is anything wrong with it???

Answer 5

a=3.7

Answer 6

rest is good:)

Answer 7

i cant believe I made this mistake and asked you.....i m such an idiot...godddd
sorryyyyy...

Answer 8

not at all...! nice work, just a tiny blip :)

Answer 9

can i ask you a chemistry doubt??

Answer 10

here's a little quicker way for future reference though
\[V _{f}^{2} =V _{i}^{2} -2ah\]
\[V _{f}^{2} =0\]
\[-V _{i}^{2} = -2ah\]
\[\frac{V _{i}^{2} }{ 2a } = h\]

Answer 11

my chemistry is pretty rusty...

Answer 12

ohhk....thank u

Answer 13

mean the equations.....