How can I use the range equation to find the initial velocity of a projectile?

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To find the initial velocity of a projectile, the range equation can be applied alongside component analysis. The hang time of 4.6 seconds provides the vertical motion information, while the horizontal distance of 23 meters is used to calculate the horizontal velocity (Vx). The equation Dy(t) = -0.5at^2 + Vit + y0 helps determine the vertical component of the velocity (Vy). By solving for Vx using the horizontal distance and time, and then calculating the resultant velocity with the equation sqrt(Vx^2 + Vy^2), the initial velocity can be accurately found. This method effectively combines both horizontal and vertical motion to arrive at the solution.
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1. A football is kicked so that is had a hang time for 4.6 s and hits the ground 23m down the field ?
What was the velocity of the football when it was kicked?
2. I used the Dx=Vx(t), don't know what to use after that.
3. Can someone show step by step how to get this, I'm having trouble trying to get an answer.
Can anyone just show me the equation I would use after that. I know I can find the xi and xf , but how do I find the yi or yf?
 
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Think about this problem in terms of components:

hang time is vertical, distance is horizontal. you should also used the equation Dy(t) = -.5at^2 + Vit + y0. from there, you can get the y component, then just find sqrt(Vx^2 + Vy^2)
 
Well this is how I set it up, similar to yours : dy= vi(t) + 0.5(a)(t)^2

I did that have got this: dy= vi(4.6) + 103.684
I didn't take into consideration that 9.8 is negative so sorry
But where do I go from there, I don't know how to solve the rest
 
You know the x distance: x0 = 0, xf = 23m, so 23 = Vx * t gives you Vx
you have the y time: solve for Vy with 0 = Dy(t). there will be 2 answers: 0 and the Vy you are looking for
 
Yeah, I know about competent breakdown. I got like 22.9 m/s for the Viy.I tested it using the range equation and it worked well. Thank you so much , it really did help me !
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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