Answer:
t = 1.099 s
Explanation:
given,
constant speed = 2.51 m/s
height of balloon above ground = 3.16 m
time elapsed before it hit the ground = ?
Applying equation of motion to the compass
[tex]y = u t + \dfrac{1}{2}at^2[/tex]
[tex]-3.16 = 2.51 t + \dfrac{1}{2}\times (-9.8)t^2[/tex]
[tex]4.9 t^2 - 2.51 t - 3.16 = 0[/tex]
using quadratic formula to solve the equation
[tex]t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]t = \dfrac{-(-2.51)\pm \sqrt{2.51^2-4(4.9)(-3.16)}}{2\times 4.9}[/tex]
t = 1.099 s, -0.586 s
hence, the time elapses before the compass hit the ground is equal to 1.099 s.