help me answer this physics question (1 Viewer)

[3unitgang]

A projectile has a range of 1240 metres and a flight time of 5 seconds. At what velocity is it projected?

 

[jazz519]

delta x = Ux * t
1240 = Ux *(5)
Ux = 248 m/s

delta y = Uy*t - 1/2 *gt^2
0 = 5Uy - 1/2 * 9.8*25
Uy=24.5 m/s

So U = sqrt (Uy^2 + Ux^2)
U = 249.2072.. m/s
Find the angle use tantheta

 

[3unitgang]

isn't the formula delta y=Uy*t + 1/2 *gt^2 and the answer you get for Uy=-24.5m/s which we in physics we can just change the negative when we want so the answer is Uy=24.5 m/s. Also, instead of using the displacement formula, can i use v=u+at where v=o, u=?, a=9.8 and t=5 to find uy?

 

[3unitgang]

Sorry, I didn't see you subbed in gravity -9.8, I thought u subbed in 9.8 sorry. However, am I able to use v=u+at to find uy?

 

[Sp3ctre]

Sorry, I didn't see you subbed in gravity -9.8, I thought u subbed in 9.8 sorry. However, am I able to use v=u+at to find uy?

No, you can't treat Vx or Vy as 0, final velocity refers to its velocity before hitting the ground, which will probably never be 0 unless the projectile is falling vertically downwards, in which case Vx would be 0.