an object is launched vertically upward from a bridge that is 30.0 m above a lak
ID: 2167420 • Letter: A
Question
an object is launched vertically upward from a bridge that is 30.0 m above a lake. The initial speed of the object is 35.0 m/s.Find:
The time it takes to reach its maximum height (=3.57s)
The maximum height above the lake (=92.5m)
The velocity of the object when it is back to its original height, where it just misses the bridge (?? not following what they want, but wouldn't it be 0m/s??)
The velocity of the object when it hits the lake (=28.77m/s I believe)
The total time of flight?
I need better understanding of which formulas to use and what data to insert where... Anything in () is an answer I derived.
Thank You In Advance!!
Explanation / Answer
A) We know the acceleration of the object is -9.8 m/s^s The equation for velocity is Vf = Vi + At When the object is at its highest point, the velocity is 0, so: 0 = 35 + (-9.8) t t = 35/9.8 = 3.57 s B) Use the equation Yf = Yi + Vit + 1/2At^2 with t=3.57 Yf = 30 +35*3.57 -4.9*(3.57)^2 = 92.50 m C) Use the equation Yf = Yi + Vit + 1/2At^2 to solve for when the height is 30 m. 30 = 30 + 35t - 4.9t^2 0 = t(35 - 4.9t) t=0 or t = 35/4.9 = 7.143 s Now use that t in the Equation Vf = Vi - At Vf = 35 - 9.8(7.143) = -35 m/s (You could also just remember that it will take the same amount of time to go up as it does down, and the the magnitude of velocity will be the same on the way up as on the way down) D) Use the equation Yf = Yi + Vit + 1/2At^2 to solve for when the height is 0 m. 0 = 30 + 35t - 4.9t^2 the solutions to this are: t= -.7734 and t=7.9163 The positive one is the only one we care about. We plug that into Vf = Vi +At Vf = 35 - 9.8 (7.9163) = -42.58 m/s E) I actually did this to solve part (D) Use the equation Yf = Yi + Vit + 1/2At^2 to solve for when the height is 0 m. 0 = 30 + 35t - 4.9t^2 the solutions to this are: t= -.7734 and t=7.9163 The positive one is the only one we care about. t = 7.92 s
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