Answer:
v = -431.2 m/s
Explanation:
Given that,
Initial position of the object, [tex]x=640-4.9t^2[/tex]
Let v is its speed 44 second after it is dropped. The relation between the speed and the position is given by :
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]v=\dfrac{d(640-4.9t^2)}{dt}[/tex]
[tex]v=-9.8t[/tex]
Put t = 44 seconds in above equation. So,
[tex]v=-9.8\times 44[/tex]
v = -431.2 m/s
So, the speed of the ball 44 seconds after it is dropped is 431.2 m/s and it is in moving downwards.
