Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
physics qs help please?
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 49.0m/s2 . The acceleration period lasts for time 8.00s until the fuel is exhausted. After that, the rocket is in free fall.
Find the maximum height ymax reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 m/s2 .
2 Answers
- 7 years agoFavorite Answer
u = 0 m/s, a = 49 m/s^2, t = 8 sec, v = ?
Using 1st equation of motion,
v = u + at
v = 0 + 49 x 8 = 392 m/s
Now you have to find distance (s)
Using 3rd equation of motion,
v^2 = u^2 + 2as
(392)^2 = 0 + 2 x 49 x s
153664 = 98s
s = (153664)/(98) = 1568 m
There is no need for acceleration due to gravity because you've asked to find only the distance.
- 7 years ago
find distance after 8secs
using
s = ut + 1/2at^2
where s is the distance you want, u = 0, a = 49.0 and t = 8sec
solve to get s = distance
A complete solution has been posted here,
