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

Answer:

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

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

And we can use the quadratic formula to solve it:

$t = \frac{2 \pm \sqrt{(-2)^2 -4(4.9)(-10)}}{2*4.9}$

$t =\frac{2 \pm \sqrt{200}}{9.8}$

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

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

Step-by-step explanation:

For this case we can use the following kinematics formulas:

$y_f = y_o + V_o t + \frac{1}{2} a t^2$

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:

$-10 m = 0m + (0m/s)t -\frac{1}{2} (9.8 m/s^2) t^2$

And solving for t we got:

$t = \sqrt{\frac{2 *10m}{9.8 m/s^2}}=1.43s$

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

$-10 m = 0m + (2m/s)t -\frac{1}{2} (9.8 m/s^2) t^2$

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:

$t = \frac{2 \pm \sqrt{(-2)^2 -4(4.9)(-10)}}{2*4.9}$

$t =\frac{2 \pm \sqrt{200}}{9.8}$

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

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