Question

Estimate the time it takes for a free fall drop from 10 meters height. Also estimate the time a 10 m platform diver would be in the air if he takes off straight up with a vertical speed of 2 m/s (and clears the platform of course!)

Answer 1

a)
`t = \sqrt{(2 *10m)/(9.8 m/s^2)}=1.43s`

b)
`4.9t^2 -2t -10 =0`

And we can use the quadratic formula to solve it:

`t = (2 \pm √((-2)^2 -4(4.9)(-10)))/(2*4.9)`

`t =(2 \pm √(200))/(9.8)`

`t_1 =1.65 s, t_2 =-1.24 s`

And since the time can't be negative the correct option would be
`t=1.65 s`

Explanation:

For this case we can use the following kinematics formulas:

`y_f = y_o + V_o t + (1)/(2) a t^2`

For this case we assume that the only acceleration is the gravity a = g =9.8 m/s^2. And for this case we can assume that the reference point is
`y_o =0` and the final height would be
`y_f = -10 m` since is below the initial point.

Teh acceleration would be a=-g, since the gravity is acting dowward, we assume that the initial velocity is 0, so then we have everything to replace and we got:

`-10 m = 0m + (0m/s)t -(1)/(2) (9.8 m/s^2) t^2`

And solving for t we got:

`t = \sqrt{(2 *10m)/(9.8 m/s^2)}=1.43s`

For the second part assuming that We have an initial vertical speed of
`v_o = 2 m/s` we have the following equation:

`-10 m = 0m + (2m/s)t -(1)/(2) (9.8 m/s^2) t^2`

And we have this quadratic equation:

`-10 = 2t -4.9t^2`

`4.9t^2 -2t -10 =0`

And we can use the quadratic formula to solve it:

`t = (2 \pm √((-2)^2 -4(4.9)(-10)))/(2*4.9)`

`t =(2 \pm √(200))/(9.8)`

`t_1 =1.65 s, t_2 =-1.24 s`

And since the time can't be negative the correct option would be
`t=1.65 s`