Question

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.

Answer 1

v = -431.2 m/s

Step-by-step explanation:

Given that,

Initial position of the object,
`x=640-4.9t^2`

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

`v=(dx)/(dt)`

`v=(d(640-4.9t^2))/(dt)`

`v=-9.8t`

Put t = 44 seconds in above equation. So,

`v=-9.8* 44`

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.