Question

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?

Answer 1

`t_+ =2.1s`
`t_- =0.5s`
#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*a_y` ; `F_x=0`
`-g=a_y` ; `a_x=0`
`v_y=v_(y0)-g*t` ; `v_x=v_(x0)=0`
`h=v_(y0)*t-g*t^2/2`

The maximum high will be when `v_y=0 => t_h=v_(yo)/g=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_+ = 1.3s + 0.8s =2.1s`
`t_- = 1.3s - 0.8s =0.5s`