Question

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)= ________

Answer 1

V(t) = 58-9.8t

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

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

Answer 2

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 :

`a(t) = -9.8 (\rm m)/(\mathrm s^2)`

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

`\displaystyle v(t) = v(0) + \int_0^t a(u) \, du`

Compute the integral:

`\displaystyle v(t) = 58 (\rm m)/(\rm s) - \left(9.8 (\rm m)/(\mathrm s^2)\right) \int_0^t du`

`\displaystyle v(t) = 58 (\rm m)/(\rm s) - \left(9.8 (\rm m)/(\mathrm s^2)\right) (t - 0)`

`\displaystyle \boxed{v(t) = 58 (\rm m)/(\rm s) - \left(9.8 (\rm m)/(\mathrm s^2)\right) t}`

Similarly, the height function is given by

`\displaystyle h(t) = h(0) + \int_0^t v(u) \, du`

`\displaystyle h(t) = 1.2 \, \mathrm m + \int_0^t \left(58 (\rm m)/(\rm s) - \left(9.8 (\rm m)/(\mathrm s^2)\right) t\right) \, du`

`\displaystyle h(t) = 1.2 \, \mathrm m + \left(58 (\rm m)/(\rm s)\right) (t - 0) - \frac12 \left(9.8 (\rm m)/(\mathrm s^2)\right) (t^2 - 0^2)`

`\displaystyle \boxed{h(t) = 1.2 \, \mathrm m + \left(58 (\rm m)/(\rm s)\right) t - \frac12 \left(9.8 (\rm m)/(\mathrm s^2)\right) t^2}`