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velikii [3]
3 years ago
10

The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob

ject will be dropped with no initial speed. The objects are initially 179 cm above the ground and separated by 142 cm. What will be the difference in the landing locations of the two objects?
Physics
1 answer:
Crazy boy [7]3 years ago
4 0

Answer:

104.3 cm  or 179.7

Explanation:

First find time that it takes for the object to hit the ground

\sqrt{(2H)/g}  ->   \sqrt{(2 x 179)/ 9.8} = 6.04s\\*

Then find xf of projectile xf= 5.9(6.04) = 37.7\\\\

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

hope that helps

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7 0
3 years ago
In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 s to attain
Sav [38]

Answer:

Laura is ahead and for a distance of 3.22 m

Explanation:

To solve this problem of one-dimensional kinematics, we have to find the acceleration and the final speed of each runner. Let's start with Laura

Lura data is acceleration time 1.82s, total run time 10.4 s and total distance 100m.  In all the races the  rest starts, so the initial speed is zero (Vo = 0)

   Vf1= Vo + a1 t1    

   Vf1 = x/t                

   XT  = X1 + X2

   X1 = Vo t1 + ½ a1 t1²  

   X1 = ½ a1 t1²  

   X2 = Vf1 (t-t1)

This is the remaining time of the race after the acceleration is over.

    XT = ½ a1 t1² + Vf1 (t-t1)

We remplace the expression of Vf1

     XT = ½ a1 t1² + a1 t1 (t-t1)

Laura's aceleration (a1) is

   a1= XT / [ ½  t1² + t1 (t-t1)]

   a1= 100/ [ ½ 1.82²+ 1.82 (10.4 -1.82)]

   a1=  5.79m/s2  

We repeat the same calculation for the other Healan runner, whose data are: total distance 100m, acceleration time 3.07 s and total time 10.4 s

Vf2= Vo + a2 t2    

Vf2 = x/t                

XT  = X3 + X4

X3 = Vo t2 + ½ a2 t2²  

X3 = ½ a2 t2²  

X4 = Vf2 (t-t2)

XT = ½ a2 t2² + Vf2 (t- t2)

XT = ½ a2 t2² + a2 t2 (t-t2)

The aceleration of Healan (a2)

a2 = XT / [½ t2² + t2 (t-t2)]

a2 = 100 / [½ 3,07²+ 3.07 (10.4 -3.07)]

a2 = 3.67 m / s2

We also need the final speeds of each runner

Laura Vf1 = Vo + a1 t1

          Vf1 = 0 + 5.79 1.82

          Vf1 = 10.54 m / s

Healan Vf2 = Vo + a2 t2

            Vf2 = 0 + 3.67 3.07

            Vf2 = 11.27 m / s

Having the acceleration and speed of each runner, you can start answering the questions

a) For t3 = 6.15s

Laura

The time to stop with constant speed is what remains after accelerating

XL= ½ a1 t1² + Vf1 (t3-t1)

XL= ½ 5.79 1.82² + 10.54 (6.15 – 1.82)    

XL= 55.23 m

Healan  

XH= ½ a2 t2² + Vf2 (t3-t2)

             XH= ½  3.67 3.07² + 11.27 (6.15-3.07)

             XH= 52.01 m

             (XL -XH)= 55.23- 52.01

             (XH -XL)=  3.22 m

It is appreciated from these results that Laura is ahead and for a distance of 3.22 m

b) If we analyze the acceleration values ​​of each runner, knowing that they leave the rest and that Healan at the end has a speed greater than Laura, the point of maximum distance difference is when Laura stops accelerating t = 1.82 s

      XL= ½  a1 t12  

      XL= ½ 5.79 1.822

      XL= 9.59 m

      XH = ½ a2 t12

      XH= ½ 3.67 1.822

      XH= 6.08 m

The maximum distance difference is 3.51 m

c) Already analyzed in the previous part 1.82 s, since the Laura stop accelerating and Heala continue with acceleration will travel greater distances in equal time units

6 0
3 years ago
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