Answer: The speed of the ball 0.70 seconds after release is 7.36 m/s
Explanation:
Using equation of motion in vertical direction and considering downward direction as positive
[tex]v_y=u_y +a_yt[/tex]
where [tex]u_y= 0.50 \frac{m}{s} , a_y=g=9.80 \frac{m}{s^{2}} , t=0.70s[/tex]
=> [tex]v_y=(0.5+9.8\times 0.70) \frac{m}{s}= 7.36\frac{m}{s}[/tex]
Thus the speed of the ball 0.70 seconds after release is 7.36 m/s
The speed of the ball at the given time after it was released is 7.36m/s
Given the data in the question
To get the speed of the ball 0.70 seconds after it is released.
We use the First Equation of Motion:
[tex]v = u + at[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration.
Since the ball was thrown from a particular height, it is under gravity and acceleration due to gravity [tex]g = 9.8m/s^2[/tex]
Hence, the equation becomes
[tex]v = u + gt[/tex]
We substitute our given values into the equation and find "v"
[tex]v = 0.50m/s + ( 9.8m/s^2\ *\ 0.70s )\\\\v = 0.5m/s + 6.86 m/s\\\\v = 7.36m/s[/tex]
Therefore, the speed of the ball at the given time after it was released is 7.36m/s
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