Use the definite integral and the Fundamental Theorem of Calculus to solve the following problem. A ball is thrown upward from a height of 1.2 m at an initial speed of 58 m/sec. Acceleration resulting from gravity is -9.8 m/sec^2. Find the following: v(t)=
h(t)=

If v(t) and h(t) denote the ball's velocity and height at time t, respectively, then we have h(0) = 1.2 m and v(0) = 58 m/s.

The ball is in free fall after being tossed, so its acceleration is constant at any time t :

By the fundamental theorem of calculus, the velocity function is given by

Compute the integral:

Similarly, the height function is given by

V(t) = 58-9.8t

h(t)=1.2+58t-4.9

Step-by-step explanation:

Using the standard formula

a=V f-Vi/t --rearrange

at=V f-Vi

V f=V i + a t

therefore

V(t)=58-9.8t

same with height

H= Hi+Vit+1/2at^2

therefore

h(t)=1.2+58t-4.9t^2

step-by-step explanation:

step-by-step explanation: