Question

I have a calculus question about initial velocity, pic included.

Answer 1

The constant acceleration of a ball shot up from the ground is a = -32 ft/sec^2.

The acceleration is defined as the instant rate of change of the velocity:

$a=\frac{dv}{dt}$

Integrating this equation, we have:

$v=\int a\cdot dt$

Since the acceleration is a constant function:

$\begin{gathered} v=\int-32\cdot dt \\ \\ v=-32\int dt \\ \\ v=-32t+v_o \end{gathered}$

The initial velocity is 48 ft/s, thus

$v=48-32t$

The ball will stop and return to the ground when the velocity is 0:

$48-32t=0$

Solving for t:

$t=\frac{48}{32}=1.5$

Now we find the displacement function by integrating the velocity:

$d=\int(48-32t)dt$

Integrating:

$\begin{gathered} d=48t-\frac{32t^2}{2}+d_o \\ \\ d=48t-16t^2+d_o \end{gathered}$

The ball was shot from the ground, so do = 0:

$d=48t-16t^2$

When t = 0, the position is d = 0.

When t = 1.5 seconds, the position is:

$\begin{gathered} d=48\cdot1.5-16(1.5)^2 \\ \\ d=72-36 \\ \\ d=36 \end{gathered}$

The ball goes up to 36 feet