Im going with A.Trade-in value
Hope this helps :)
Answer:
acute
Step-by-step explanation:
First note down the relevant variables from the question.
Ua (Initial velocity a) = 320ft/s
Ub (initial velocity b) = 240ft/s
Aay (acceleration of a in the vertical axis) = Aby = -32.17ft/s/s
We want to know when they will be at the same height so should use the formula for displacement:
s = ut + 1/2 * at^2
We want to find when both firework a and firework b will be at the same height. Therefore mathematically when: say = sby (the vertical displacements of firework A and B are equal). We also know that firework B was launched 0.25s before firework A so we should either add 0.25s to the time variable for the displacement formula for firework B or subtract 0.25s for firework A.
SO:
Say = Sby
320t + 1/2*-32.17t^2 = 240(t+0.25) + 1/2 * -32.17(t+0.25)^2
320t - 16.085t^2 = 240t + 60 - 16.085(t+0.25)^2
320t - 16.085t^2 = 240t + 60 - 16.085(t^2 + 0.5t + 6.25)
320t - 16.085t^2 = 240t + 60 -16.085t^2 - 8.0425t - 100.53
320t - 240t - 8.0425t - 16.085t^2 + 16.085t^2 = 60 - 100.53
71.958t = -40.53
t = -0.56s (negative because we set t before Firework A was launched)
Now we know both fireworks explode 0.56 seconds AFTER fireworks B launches (because we added 0.25 seconds to the t variable in the equation above for the vertical displacement of Firework B).
You could continue on to find the displacement they both explode at and verify the answer by ensuring that it is equal (because the question stated they should explode at the same height by substituting the value we found for t of 0.56s into the vertical displacement formula for firework A and t+0.25s=0.81s into the same formula for Firework B
Verification:
Say = ut + 1/2at^2
Say = 320*0.56 + 1/2*-32.17*0.56^2
Say = 179.2 + -5.04
Say = 174.16ft
Sby = ut + 1/2at^2
Sby = 240*0.81 + 1/2*-32.17*0.81^2
Sby = 194.4 - 10.5
Sby = 183.9ft
While Say is close to Sby I would have expected them to be almost perfectly equal… can you please check if this matches the answer in your textbook? There could be wires due to rounding. I also usually work in SI units which use the metric system and not the imperial system although that shouldn’t make a difference. The working out and thought process is correct though and this is why trying to verify the answer is an important step to make sure it works out.
Answer: 0.56s (I think)
I would say A. would be your answer
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the following Sum to Product Identities:
<u>Proof LHS → RHS</u>
![\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cqquad%20%5Cdfrac%7B%5Ccos%2010%5Cbigg%5B2%5Ccos%20%5Cbigg%28%5Cdfrac%7B15%2B25%7D%7B2%7D%5Cbigg%29%5Csin%20%5Cbigg%28%5Cdfrac%7B15-25%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%7D%7B%5Ccos%2020%5Cbigg%5B-2%5Csin%20%5Cbigg%28%5Cdfrac%7B15%2B5%7D%7B2%7D%5Cbigg%29%5Csin%20%5Cbigg%28%5Cdfrac%7B15-5%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%7D)
![\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10](https://tex.z-dn.net/?f=%5Ctext%7BSimplify%3A%7D%5Cqquad%20%5Cqquad%20%5Cdfrac%7B%5Ccos%2010%5B2%5Ccos%2020%5Csin%20%28-5%29%5D%7D%7B%5Ccos%2020%5B-2%5Csin%2010%5Csin%205%5D%7D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D%5Cdfrac%7B-2%5Ccos10%20%5Ccos%2020%20%5Csin%205%7D%7B-2%5Csin%2010%20%5Ccos%2020%20%5Csin%205%7D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D%5Cdfrac%7B%5Ccos%2010%7D%7B%5Csin%2010%7D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D%5Ccot%2010)
LHS = RHS: cot 10 = cot 10 
