Catch This Embarrassing Moment on Camera

$\begin{cases} \alpha = - 9.8 \; m/s^2 \\ V_o = 0 \; m/s \\ S_o = 10 \; m \\ S = 5 \; m \\ t = \end{cases} \\\\ S = S_o + V_o \cdot t + \frac{\alpha \cdot t^2}{2} \\\\ 5 = 10 + 0 + \frac{- 9.8t^2}{2} \\\\ \frac{9.8t^2}{2} = 5 \\\\ 9.8t^2 = 10 \\\\ t^2 = 1,020 \\\\ \boxed{t = 1.01 \; s}$

$t = \sqrt{\frac{2h}{g}}$ which h = 10/2 = 5 meters

H = (g.t^2)/2

5 = 4,9t^2 t = 1,01

To solve this problem, we use the equation $d=d_0+vt+\frac{1}{2}at^2$ . Note that in this particular problem, we have that $v=0,a=9.8$ . We set $d_0=0$ for simplicity. We are trying to solve for the time $t$ when the teenager has gone $5$ meters. Thus the equation is $5=\frac{1}{2}9.8t^2$ or $t=\sqrt{\frac{10}{9.8}}\approx\boxed{1.010}$

=h/g =10/9.8 =1.020

teenafer felt half way that is s2-s1=10/2=5m s2-s1=ut+1/2at^2 u=0m/s a=g

solve 5=0+1/2 10 t^2 t^2=1 t=1s(negative value not possible)

Use II Eq. of Motion i.e. s=u t+1/2 g*t^{2} u=0 and s=5m(half way down) then after solving t=1.01 sec

Given, h=10 m (halfway height= 5 m) u=0 m/s (The boy was initially at rest) g=9.8 m/s ^2

we know, by the second equation of motion that, h=u t+1/2 at^2

put the values and you will get the correct answer. (Put h=5 m)

Hello,y axis motion...

halfway down=5m,as initial velocity=0 m/s, so by applying v^2=u^2-2gs,substitute u=0 m/s and g= -9.8m/s/s, V^2=0^2-(2x-9.8x5)=98 therefore,v=(98)^0.5,use v=u-gt to find the time,so t=v-u/-g=(98)^0.5 - 0 / (-(-9.8))=~1.01s..

thanks...

Half of 10m = 5m, so : 5 = 10 + 0.t - 4.9.t.t => 0=5 - 4.9 t^2 => t^2 = 1.001 => t = 1.01s

5=0-0.5 9.8 t^2 solve it. you will get t=1

Using S=So +vot + at²/2:

5 = 10t²/2

t = 1s

H = g . t^2 /2 5 = 9,8 . t^2 /2 5 = 4,9 . t^2 t^2 = 1,02 t = 1,01