intro to physics
posted by emilia on .
a toy rocket moving vertically upward passes by a 2.2m high window whose sill is 7.0m above the ground. the rocket takes 0.17s to travel the 2.2m height of the wondow
part a): what was the launch speed of the rocket ? assume the propellant is burned very quickly at blastoff
part b) how high will the rocket go ?

Let’s find the initial velocity of rocket.
Since
h(o) = v(o)•t  g•t^2/2,
v(o) = 2• h(o) /t +gt/2 = 22.2/0.17 + 9.8•0.17/2 = 25.047 m/s.
Now let’s find the time of reaching the max height
0 = v(o)  g•t1
t1 = v(o)/g = 25.047/9.8 = 2.56 s.
The max height (from the sill)
h1 = v(o)•t1gt1^2/2 = 25.047•2.56 =9.8•(2.56)^2/2 = 96.13 m.
The max height (from the ground) is
H = 96.13 + 7 = 103.13 m.
H = gt2^2/2 .
t2= sqrt(2•H/g) = sqrt(2•103.13/9.8) = 4.59 s.
v =g•t2 = 9.8•4.59 = 44.96 m/s 
where did that 22.2 come from?

v(o) = 2• h(o) /t +gt/2 = 2.2/0.17 + 9.8•0.17/2 = 25.047 m/s.

22.2 was a typo error of 2*2.2, but the result was ok.

thats wrong, no 2* h(o)