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Sorin hits a cricket ball straight up into the air. After 1.5\,\text{s}1.5s1, point, 5, start text, s, end text, the ball is falling straight down with a speed of 6.5\,\dfrac{\text{m}}{\text{s}}6.5
s
m
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6, point, 5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. We want to find the initial vertical velocity of the ball. We can ignore air resistance.

Respuesta :

skyluke89

Answer:

8.2 m/s upward

Explanation:

The motion of the ball is a free fall motion, so we can use the suvat equation:

[tex]v=u+at[/tex]

where:

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

Choosing upward as positive direction:

[tex]v=-6.5 m/s[/tex]

[tex]a=g=-9.8 m/s^2[/tex] (acceleration of gravity)

t = 1.5 s

Solving for u, we find the initial velocity:

[tex]u=v-at=-6.5-(-9.8)(1.5)=8.2 m/s[/tex]

And the direction is upward.