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konstantin123 [22]
3 years ago
9

How long will it take a projectile to hit the ground if it is launched at 4.2 m/s off a 12.6 m high platform?

Physics
1 answer:
8090 [49]3 years ago
6 0

Answer: 3

Explanation:

I think you are supposed to divided in those problems? I'm not sure but if you aren't then please let me know. The equation I did was:

12.6 / 4.2 = 3

Hope this helps!! :)

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A body has a mass of 2kg.it accelerats from 20m/s to 40m/s in 4 seconds.the resultant force is<br>​
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The resultant force is 8N  

Given that mass is 2kg , v= 40m/s, u =20m/s and we need to calculate resultant force
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The difference between the forces that are acting on an object as part of a system is known as the resultant force.
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Sam lifted his backpack with 5 Newtons of force a total of 400
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3 years ago
An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's a
devlian [24]

Answer:

a

D =  1162.7 \  m

b

\beta =- 65.55^o

Explanation:

From the question we are told that

  The speed of the airplane is  u  =  92.3 \ m/s

   The  angle is  \theta = 51.1^o

    The altitude of the plane is  d =  532 \  m

Generally the y-component of the airplanes velocity is  

       u_y  =  v *  sin (\theta )

=>     u_y  =   92.3 *  sin ( 51.1 )

=>     u_y  =  71.83  \ m/s

Generally the displacement  traveled by the package in the vertical direction is

       d =  (u_y)t +  \frac{1}{2}(-g)t^2

=>       -532  = 71.83 t +  \frac{1}{2}(-9.8)t^2

Here the negative sign for the distance show that the direction is along the negative y-axis

 =>   4.9t^2 - 71.83t - 532 = 0

Solving this using quadratic formula we obtain that

    t =  20.06 \  s

Generally the x-component of the velocity is  

     u_x  =  u  *  cos (\theta)

=>    u_x  =   92.3  *  cos (51.1)

=>   u_x  =   57.96 \ m/s

Generally the distance travel in the horizontal  direction is    

     D =  u_x  *  t

=>   D =  57.96  *   20.06

=>    D =  1162.7 \  m

Generally the angle of the velocity vector relative to the ground is mathematically represented as

       \beta  =  tan ^{-1}[\frac{v_y}{v_x } ]

Here v_y is the final  velocity of the package along the vertical  axis and this is mathematically represented as  

     v_y  =  u_y  -   gt

=>  v_y  =  71.83  -    9.8 *  20.06

=>  v_y  =  -130.05 \  m/s  

and  v_x is the final  velocity of the package which is equivalent to the initial velocity u_x

So

       \beta  =  tan ^{-1}[-130.05}{57.96 } ]

       \beta =- 65.55^o

The negative direction show that it is moving towards the south east direction

   

6 0
3 years ago
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