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?

-9.8 m/s^2

To find gravity's constant value, we need to understand the relationship between the position, velocity, and acceleration of the object in free fall.

The position of the object is given by the function d(t) = -4.9t^2 + 50, where d is the distance from the ground and t is the time in seconds.

The velocity of the object is the derivative of the position function. Taking the derivative of d(t) with respect to t, we get:

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

The negative sign indicates that the velocity is decreasing over time.

Acceleration, denoted as a, is the derivative of velocity with respect to time. Taking the derivative of v(t) with respect to t, we get:

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

Therefore, the constant value of gravity is -9.8 m/s^2.

Note that in this case, we assumed that the positive direction is upwards. If the positive direction is downwards, the value of gravity will be positive (+9.8 m/s^2).