contestada

A ball is launched with an initial horizontal velocity of 10.0 meters per second. It takes 500 milliseconds for the ball to reach its maximum height.
Determine the maximum horizontal distance that the ball will travel.
Calculate the initial vertical velocity of the ball.

Respuesta :

Answer:

500ms times 2 would be when the ball reaches the max horizontal distance.

Then to find the angle, use the formula of time to reach max height t = u sin theta / g . With t being the max height time 500ms, u being 10m/s

For initial vertical velocity just use u sin theta.

Cricetus

The max horizontal height is "10 meters" and the initial vertical velocity is "4.9 m/s".

Given:

Horizontal velocity,

  • [tex]V_x = 10 \ m/s[/tex]

Time,

  • t = 500 m/s

At max height,

  • [tex]V_{yf} = 0 \ m/s[/tex]

(a)

→ Time to flight (T) will be:

= [tex]2t[/tex]

= [tex]2\times 0.5[/tex]

= [tex]1 \ second[/tex]

→ Horizontal distance will be:

= [tex]V_x\times T[/tex]

= [tex]10\times 1[/tex]

= [tex]10 \ meters[/tex]

(b)

→ The initial vertical velocity will be:

[tex]V_{yf} = V_y +gt[/tex]

   [tex]0 = V_y - 9.8\times 0.5[/tex]

  [tex]V_y = 4.9 \ m/s[/tex]

Thus the above answers are correct.

Learn more:

https://brainly.com/question/3698280

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