Respuesta :

zeshan
Using first equation of motion;
vf = vi + at --------------------- (1)
where vf = final velocity
vi = initial velocity
a = acceleration (here it is considered to be gravitational acceleration)
t = time

As
vi = 1.8 m/s
vf = 0 m/s
a = g = -9.8 m/s^2 (negative sign is due to the upward motion of tina)

using equation (1), 
0 = 1.8 + (-9.8 * t)
t = 9.8/1.8
t = 0.1836 seconds
but the tina has to travel back to the ground, hence the time taken by tina to be in the air will be 
t = 2 * 0.1836

t = 0.367 seconds
skyluke89

Answer:

0.367 s

Explanation:

Tina has a unifomrly accelerated motion along the vertical direction, with initial velocity [tex]v_0=1.8 m/s[/tex] and constant acceleration [tex]g=-9.8 m/s^2[/tex] downward (acceleration due to gravity). Its position at time t is given by the equation:

[tex]y(t)=y_0 +v_0t +\frac{1}{2}gt^2[/tex]

where [tex]y_0=0[/tex] is the initial height, so we can remove it from the equation:

[tex]y(t)=v_0 t + \frac{1}{2}gt^2[/tex]

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 [tex]y(t)=0[/tex]:

[tex]0=v_0 t +\frac{1}{2}gt^2\\t(v_0+\frac{1}{2}gt)=0[/tex]

which has two solutions:

[tex]t=0 s[/tex] --> time at which Tina starts its motion, so we don't consider this one

[tex]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[/tex]

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