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anonymous
 one year ago
an object is projected straight up with an initial velocity of k ft. If the object returns to the ground at t=8 sec., find the height (in feet) of the object at t=7 sec
anonymous
 one year ago
an object is projected straight up with an initial velocity of k ft. If the object returns to the ground at t=8 sec., find the height (in feet) of the object at t=7 sec

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whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0The general equation is\[h(t) = \frac{1}{2}gt^2 + v_0 t + h_0\]where \(g\) is the acceleration due to gravity, \(v_0\) is the initial velocity (\(k\) in this problem), and \(h_0\) is the initial height (presumably \(h_0=0\) as unspecified in this problem). At any time \(t\) where \(h(t)=0\), the object is at ground level. As we are using ft as the unit of measurement, use \(g = 32.2 \text{ft/s}^2\) That gives you a quadratic to solve: \[0 = \frac{1}{2}(32.2)t^2 + k t\] Find the value of \(k\) that makes that equation true at \(t = 8\). Then plug it into your initial equation and evaluate at \(t=7\) to find the height of the object at \(t=7\) seconds.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you please explain like step by step please

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Well, I thought I did :) How about you tell me where my explanation stops making sense, and I'll go from there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so this is a project but I honestly don't understand any of this so can you break it don how you got 32 and how you plug it in

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Okay. Is this for a math class or a physics class? I'm guessing math...

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Fair enough. Well, the first thing you need to know is that if you ignore air resistance, the path of an object thrown up in the air is a parabola. Do you know the general equation for a parabola?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0h(t)=−12 gt 2 +v 0 t+h 0 im guessing it is this correct?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0A parabola can be expressed as \[ax^2+bx+c\]where \(a,b,c\) are all constants I've used different names for those constants to make more physical sense, but the form is the same.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Now, if the parabola represents something thrown up in the air (without sufficient velocity to escape the pull of gravity), the path will look something like this: dw:1443649458279:dw Agreed?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you for working with us, can you do a few more?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0The equation I initially gave is the form that a physicist would use. It has 3 components to determine the position/height of the object: \[\frac{1}{2}gt^2\]represents the contribution from gravity. This pulls the object down, and gets larger and larger as time goes on (constant acceleration for a longer time means more movement). If you just had an object dropped from the top of a building with no initial velocity, this term alone would give you position relative to the starting point. \[v_0 t\]represents the contribution from the initial velocity upward. If you had no gravity or other forces acting on the thrown object, this term would give you constant motion with no acceleration. distance = speed * time \[h_0\]represents your starting height, and in our case is \(0\) If you add the three together, you get the total "picture" of the motion.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0The 32.2 ft/s^2 comes from assuming this problem takes place on the Earth's surface. That's the approximate value of the acceleration due to gravity here.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0That's how I came up with the equation. Now we need to do the work! \[h(t) = \frac{1}{2}(32.2 \text{ft/s}^2)t^2 + k t + 0\] We know that at \(t=0\), \(h(t)=0\) because we haven't thrown the ball yet \[0 = \frac{1}{2}(32.2 \text{ ft/s}^2)(0)^2 + k(0) + 0\] but that is all \(0\) for each term, so that didn't help us at all.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0We also know that the object is back on the ground at \(t = 8 \text{ s}\), so \[h(8\text{ s}) = 0 = \frac{1}{2}(32.2\text{ ft/s}^2)(8\text{ s})^ + k(8\text{ s}) + 0\] or if we remove the units for clarity: \[h(8) = 0 = \frac{1}{2}(32.2)(8)^2 + 8k\] Can you solve that last equation for \(k\)?
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