# A ball rolls horizontally off a table of height 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the ground?

Jan 19, 2017

As the ball rolls horizontally of a ta!ble (of height 1.4m) with a speed of $4 \text{m/s}$ ,the horizontal component of velocity will be ${v}_{\text{hor"=(4cos0)"m/s"=4"m/s}}$ and vertical component will be ${v}_{\text{ver"=(4cos90)=0"m/s}}$

If the time required to reach the be t sec then by the equation of motion under gravity we can write

$h = {u}_{\text{ver}} \times t + \frac{1}{2} \times g \times {t}^{2}$

Taking $\text{acceleration due to gravity }$
$g = 9.8 \text{m/} {s}^{2} \mathmr{and} h = 1.4 m$ we get

$\implies 1.4 = 0 \times t + \frac{1}{2} \times 9.8 \times {t}^{2}$

$\implies t = \sqrt{\frac{2}{7}} s$