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launched at ground level with an initial speed of 41 m/s, at an angle of 29° above the horizontal. It strikes a target above the ground 2.3 seconds

What is the vertical distance, in meters, from where the projectile was launched to where it hits the target?

Respuesta :

bharathparasad577

The horizontal distance is 82.455m and vertical distance is 19.734m

Let us first find the x and y components of velocity

The velocity is given to be 41m/s

the angle of projectile is given to be 29 degree

The x component of velocity = Vx =v cosα =41 x 0.8746 =35.85 m/s

The y component of velocity= Vy= v sinα=41 x 0.484 =19.85 m/s

The horizontal displacement x= Vxt = 35.85 x 2.3 =82.455 m

To solve for vertical distance y = Vyt+1.2at^2

y = 19.85 x 2.3 + 1/2 x -9.8 m/s^2 x 2.3

  = 19.734m

Hence, the horizontal distance is 82.455m and vertical distance is 19.734m

For further reference:

https://brainly.com/question/14375940?referrer=searchResults

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