Question

A 200 g model rocket is observed to rise 100 m above the ground after launch. What must have been the launch speed of the rocket at the ground?

Answer 1

v = 44.75 m/s

Step-by-step explanation:

You can use the Conservation of Energy formula here

K₁ + U₁ = K₂ + U₂ or mgh₁ +
`(1)/(2)`mv²₁ = mgh₂ +
`(1)/(2)`mv²₂

We can then plug in our values

(200g)(10m/s²)(0m) +
`(1)/(2)`(200g)(v²) = (200g)(10m/s²)(100m) +
`(1)/(2)`(200g)(0v²)

0 + 100v² = 200000 + 0

100v² = 200000

v² = 2000

v =
`√(2000)`

v = 44.72 m/s

Answer 2

Conservation of Energy


`Eki + Epi = Ekf + Epf\\ (1)/(2) mv^2 + 0 = 0 + mgh\\ v= √( 2gh) =√( 2(9.8)(100))\\ v=88.5 m/s`