Answer to Question #248411 in Calculus for Bman

Question

A ball tossed in the air vertically from ground level returns to earth 4 seconds later. Find the inital velocity and maximum height of the ball.

Answer 1

According to the Galilei formulas:

$s(t) =s{\scriptscriptstyle 0}+v{\scriptscriptstyle 0}*t - {\frac 1 2}gt^{2}$

$v(t) = v{\scriptscriptstyle 0}-gt$

Where $s{\scriptscriptstyle 0}$ - initial height, $v{\scriptscriptstyle 0}$ - initial speed, g - acceleration of the gravity(9.8 m/s), t - point of time

$s{\scriptscriptstyle 0} = 0 \to s(t) = v{\scriptscriptstyle 0}*t - {\frac 1 2}gt^{2}$

The graph of s(t) takes form of parabola with branches pointing down(since $-{\frac 1 2}g <0$) , which means the t-point of the vertex of the parabola is the point where value of s(t) is maximized(maximum height of the ball)

We have s(0) = s(4), which means points t = 0 and t = 4 are located on the different branch and equidistant from the axis of the symmetry of parabola(which takes form x = $t{\scriptscriptstyle m}$ , where $t{\scriptscriptstyle m}$ - t-point of the vertex of the parabola.

So, $t{\scriptscriptstyle m} = {\frac {0+4} 2} = 2$

It took 2 seconds for the ball to reach it's maximum height

$s(2) =2v{\scriptscriptstyle 0} - 2g$

To find the value of s(2) we should find $v{\scriptscriptstyle 0}$

Since s(2) is maximum, then $s'(2) = v(2) =0$

$v(2) = v{\scriptscriptstyle 0}-2g$

$v{\scriptscriptstyle 0} = 2g$

$v{\scriptscriptstyle 0} = 19.6$ m/s

So, by substituting the obtained values we get

$s(2) =2*19.6 - 2*9.8 = 19.6$ m

The initial velocity was 19.6 m/s, the maximum height was 19.6 m.