Projectile Motion (1 Viewer)

[rawker]

..it seem's a few people are having trouble with this!

How do I answer this:
A body is projected vertically upwards with an intial velocity of 50m.s^-1
a) velocity after 2s.
b) max height it reaches
c) velocity with which it returns to Earth


I though you would answer velocity as v=u-at (50-9.8x2) and max height as s=(u+1/2gt^2) (50=1/2x9.8x4)... but I got the same answer (30.4) so I figure that I am wrong... So how would I do it, and the last one.

 

[exa_boi87]

v^2 = u^2 + 2as .. imo use that, where "v = 0" (at max height), "u = 50", "a = -9.8", then "s" is what you're after .. i get 127.6m

not sure about the last point .. did this in about a minute while the OC credits were rolling , will give it a shot lata 2nite

 

[香港!]

..it seem's a few people are having trouble with this!

How do I answer this:
A body is projected vertically upwards with an intial velocity of 50m.s^-1
a) velocity after 2s.
b) max height it reaches
c) velocity with which it returns to Earth


I though you would answer velocity as v=u-at (50-9.8x2) and max height as s=(u+1/2gt^2) (50=1/2x9.8x4)... but I got the same answer (30.4) so I figure that I am wrong... So how would I do it, and the last one.

a)v=u+at=50-9.8 (2)=30.4 m\s
b)max height is when v=0
so 0=50-9.8t
t=50\9.8
so at dis time, max height, S=ut+(1\2)at²=50(50\9.8)-4.9(50\9.8)²=127.55m
c)It returns to Earth at v=50-9.8(50\9.8 x 2)=-50 m\s