# A baseball is thrown vertically with a velocity of 13 m/s. How high does it go? At what two times is the ball 5 m above the ground?

Mar 9, 2016

${t}_{+} = 2.1 s$
${t}_{-} = 0.5 s$
h_max =8.5m

#### Explanation:

Using Newton's second law F =m * a and taking in account that the ball was thrown vertically, i.e., there is not horizontal component,

${F}_{y} = m \cdot {a}_{y}$ ; ${F}_{x} = 0$
$- g = {a}_{y}$ ; ${a}_{x} = 0$
${v}_{y} = {v}_{y 0} - g \cdot t$ ; ${v}_{x} = {v}_{x 0} = 0$
$h = {v}_{y 0} \cdot t - g \cdot {t}^{2} / 2$

The maximum high will be when ${v}_{y} = 0 \implies {t}_{h} = {v}_{y o} / g = \frac{13}{10} s$
the maximum high will be h_max=13*1.3 m - 10*1.3^2/2 m =8.5m

The moments when the ball reach 5 m are

t_(+-) = v_(yo)/g +- sqrt ((v_(yo)^2/g^2) - 2*h/g

${t}_{+} = 1.3 s + 0.8 s = 2.1 s$
${t}_{-} = 1.3 s - 0.8 s = 0.5 s$