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Projectile Motion—A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0° below the horizontal. How far does the ball move horizontally before it hits the ground?

Respuesta :

rafaleo84

Answer:

The distance will be x = 41.7 [m]

Explanation:

We must first find the components in the x & y axes of the initial velocity.

[tex](v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s][/tex]

The acceleration is the gravity acceleration therefore.

g = 9.81 [m/s^2]

Now we can calculate how long it takes to fall.

[tex]y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s][/tex]

With this time we can find the horizontal distance that runs the projectile.

[tex]x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m][/tex]

boffeemadrid

The horizontal distance the ball traveled is required.

The distance the ball moved horizontally is 27.1 m.

y = Displacement in y direction = 28 m

u = Initial velocity = 15 m/s

[tex]\theta[/tex] = Angle = [tex]20^{\circ}[/tex]

[tex]a_y[/tex] = g = Acceleration in y direction = [tex]9.81\ \text{m/s}^2[/tex]

t = Time

Displacement in y direction is given by

[tex]y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 28=15\sin 20^{\circ}t+\dfrac{1}{2}\times 9.81\times t^2\\\Rightarrow 28=5.13t+4.905t^2\\\Rightarrow 4.905t^2+5.13t-28=0\\\Rightarrow 4905t^2+5130t-28000=0\\\Rightarrow t=\dfrac{-5130\pm \sqrt{5130^2-4\times 4905\left(-28000\right)}}{2\times 4905}\\\Rightarrow t=1.92,-2.96[/tex]

The time taken to reach the ground is 1.92 s

Displacement in x direction

[tex]x=u_xt\\\Rightarrow x=15\cos 20^{\circ}\times 1.92\\\Rightarrow x=27.1\ \text{m}[/tex]

The distance the ball moved horizontally is 27.1 m.

Learn more:

https://brainly.com/question/15301091?referrer=searchResults

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