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# Physics projectile motion help watch

1. A particle is projected from a point c with horizontal and upward velocity 50ms and 30ms respectively find after how long particle is at same level as c
How to get answer which is 6.12s
2. (Original post by Hasham123)
A particle is projected from a point c with horizontal and upward velocity 50ms and 30ms respectively find afterglow long particle is at same level as c
How to get answer which is 6.12s
3. (Original post by DreamlinerFinder)
Sure
Sv= UV x T + AT^2
0=50T -4.9T^2
Then put that into quadratic formula and got 10.204 which is wrong
4. (Original post by Hasham123)
Sure
Sv= UV x T + AT^2
0=50T -4.9T^2
Then put that into quadratic formula and got 10.204 which is wrong
I think it was the quadratic formula step, but I can't remember how to do this, so I'll leave this for someone who knows it better

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5. Use v = u +at for the vertical. For the time the ball is in the air is dependent only on the vertical not the horizontal.
Arrange the formula to get t = (v - u)/a
t = (30)/9.81
t=3.058 This is the time taken for the particle to reach its max height so you will have to multiply by 2.
3.058 x 2 = 6.12
6. s = ut + 1/2at^2

0 = 30t - 4.9t^2

0 = t(30 - 4.9t)

t = 0 (not the answer you're looking for)

30 - 4.9t = 0
so t = 6.12s

7. Thank you from this I got -4.9t^2 + 30t =0 then put that in quadratic formula and got 6.12

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