An object is dropped from the top of a cliff 640 meters high. its height above the ground t seconds after it is dropped is 640−4.9t^2. determine its speed 44 seconds after it is dropped.

v = -431.2 m/s

Explanation:

Given that,

Initial position of the object,

Let v is its speed 44 second after it is dropped. The relation between the speed and the position is given by :

Put t = 44 seconds in above equation. So,

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.

double the distance = divide strength by 4