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A basketball rolls onto the court with a speed of 4 . 0 m s 4.0 s m ​ 4, point, 0, space, start fraction, m, divided by, s, end fraction to the right, and slows down with a constant acceleration of 0 . 5 0 m s 2 0.50 s 2 m ​ 0, point, 50, space, start fraction, m, divided by, s, start superscript, 2, end superscript, end fraction over 1 4 m 14m14, space, m. What is the velocity of the basketball after rolling for 1 4 m 14m14, space, m?

Respuesta :

skyluke89

Answer:

1.4 m/s

Explanation:

The final velocity of the ball can be found by using the following SUVAT equation:

[tex]v^2-u^2=2ad[/tex]

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance travelled

In this problem,

u = 4.0 m/s

d = 14 m

a = -0.50 m/s^2

Solving the equation for v,

[tex]v=\sqrt{u^2+2ad}=\sqrt{4^2+2(-0.50)(14)}=1.4 m/s[/tex]

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