Question

URGENT PLEASE ANSWER ASAP!!!!!!!!!!!

An object in free fall has its distance from the ground measured by the function d(t) =–4.9t^2 + 50, where d is in meters and t is in seconds. If gravity is the only acceleration affecting the object, what is gravity’s constant value (include units in your answer)?

Answer 1

Answer: The gravity's constant value is
`9.8 m/sec^2`.

Step-by-step explanation:

The given function shows the distance from the ground (in meters) at time t (in sec).

`d(t)=-4.9t^2+50`

First derivative of displacement is velocity and the second derivative of displacement is acceleration.

Differentiate the given equation with respect to t.

`v(t)=d'(t)=-9.8t+0`

Differentiate the above equation with respect to t.

`a(t)=v'(t)=d''(t)=-9.8`

It is given that gravity is the only acceleration affecting the object and the negative sign shows the downward direction of the object.

Therefore, the gravity's constant value is
`9.8 m/sec^2`.