The new initial velocity should be 'root 2 times u'. Tell me if there's something wrong with it or you don't understand a part of it: https://imgur.com/PuPIDE5
A projectile fired at a speed of u and an angle of x to the horizontal, reaches a
maximum height of H. When the same projectile is fired again at the same
angle with a new higher initial speed it reaches a new maximum height of 2H.
What is the new initial velocity in terms of u.
HSC 2018
English Adv | MX1 | MX2 | Phys | Chem | Eco
The new initial velocity should be 'root 2 times u'. Tell me if there's something wrong with it or you don't understand a part of it: https://imgur.com/PuPIDE5
Last edited by tazhossain99; 6 Dec 2017 at 7:32 PM.
ATAR Goal: 95ish
ATAR Achieved: 97.65
2017 HSC ~ Penrith High School
English (Adv.) [83] ~ Mathematics Ext. 1 [99] ~ Mathematics Ext. 2 [92] ~ Physics [90] ~ Biology [93]
2018-2023 ~ UNSW
BE/ME [Electrical]
That's for the new uy. But for the new initial velocity, which is root of uy2 + ux2, I got u times root (sin squared x plus 1).
Not sure if it's correct, so a confirmation would be appreciated.
Oh,got it now. I thought horizontal velocity was the same, until I realised the question didn't mention that. Thanks for the help.
There are currently 1 users browsing this thread. (0 members and 1 guests)
Bookmarks