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xeze [42]
2 years ago
6

A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How lo

ng is required to reach this height?
For part A) I did: y=11m, g=9.81m/s^2, Vy0= 18 m/s, y0=0, (V^2)y=?

V^2 y= V^2 y0- 2g (y-y0)
--> V^2 y=18^2m/s-2(9.8)(11m-0m)
--> Vy=Square root ( 324m/s-215.6)= 10.4m/s

Part B) I started out with:
y=y0 + Vy0t - (1/2) gt^2
--> 11= 18t m/s-(1/2) 9.8t^2
--> -11 + 18t- 9.8t^2
Physics
1 answer:
aliina [53]2 years ago
5 0
For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative. 
So <span>Vf = square root [2(-9.8)(11.0) + (18.0)^2] </span>
<span>Vf = 10.4 m/s your answer is correct. 

For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, </span>11= 18t m/s-(1/2) 9.8t^2 then <span>-11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s</span><span>
 </span> 
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Explanation:

It is given that,

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Below is the solution:

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