Answer:
0.367 s
Explanation:
Tina has a unifomrly accelerated motion along the vertical direction, with initial velocity \(v_0=1.8 m/s\) and constant acceleration \(g=-9.8 m/s^2\) downward (acceleration due to gravity). Its position at time t is given by the equation:
\(y(t)=y_0 +v_0t +\frac{1}{2}gt^2\)
where \(y_0=0\) is the initial height, so we can remove it from the equation:
\(y(t)=v_0 t + \frac{1}{2}gt^2\)
We need to find how long is Tina in the air, which means we have to find the time t at which Tina reaches the ground again, so the time t at which \(y(t)=0\):
\(0=v_0 t +\frac{1}{2}gt^2\\t(v_0+\frac{1}{2}gt)=0\)
which has two solutions:
\(t=0 s\) --> time at which Tina starts its motion, so we don't consider this one
\(v_0 +\frac{1}{2}gt^2 =0\\t=\frac{-2v_0}{g}=\frac{-2(1.8 m/s)}{-9.8 m/s^2}=0.367 s\)
