Question

How long is Tina, a ballerina, in the air when she leaps straight up with a speed of 1.8 m/s?

Answer 1

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

$v_0 +\frac{1}{2}gt^2 =0\\t=\frac{-2v_0}{g}=\frac{-2(1.8 m/s)}{-9.8 m/s^2}=0.367 s$

Answer 2

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