The velocity of the ball as it passes a treetop 4. 88 m tall and thrown straight upward with a velocity of 9. 98 m/s is 14 m/s.
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
The third equation of the motion for distance can be given as,
[tex]v^2-u^2=2ah[/tex]
Here, is the u initial body, a is the acceleration of the body and h is the height of it.
The ball is thrown straight upward with a velocity of 9. 98 m/s. The height of it is 4. 88 m. Thus, initial velocity is 4.88 meters and the acceleration due to gravity is 9.98 m/s².
Put the values,
[tex]v^2-u^2=2ah\\v^2-(9.98)^2=2(9.98)(4.88)\\v^2=2(9.98)(4.88)+(9.98)^2\\v\approx 14\rm\; m/s[/tex]
Hence, the velocity of the ball as it passes a treetop 4. 88 m tall and thrown straight upward with a velocity of 9. 98 m/s is 14 m/s.
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