Question #6f91b

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Question #6f91b

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Answer 1 · 2015-07-01T15:46:24

I found the following values (in red in the explanation):
(but check carefully my math!!!)

Explanation:

Have a look:
enter image source here

Answer 2 · 2015-07-02T13:23:35

Have a look if it is clearer.

Explanation:

Using the same diagram.
Considering that:
$v_{i} = 7.25 \frac{m}{s}$ $v_i=7.25 m/s$
$h_{1} + h_{2} = 25 m$ $h_1+h_2=25m$
Use: $v_{f} - v_{i} = a t$ $v_f-v_i=at$
$0 = 7.25 - 9.81 t_{1}$ $0=7.25-9.81t_1$
$t_{1} = 0.74 s$ $t_1=0.74s$
Use:
$h_{1} = v_{i} t_{1} + \frac{1}{2} a t_{1}^{2}$ $h_1=v_it_1+1/2at_1^2$
$h_{1} = 7.25 \cdot 0.74 - \frac{1}{289.81} {(0.74)}^{2}$ $h_1=7.25*0.74-1/289.81(0.74)^2$
$h_{1} = 2.68 m = 2.7 m$ $h_1=2.68m=2.7m$

$h_{2} = 25 - 2.7 = 22.3 m$ $h_2=25-2.7=22.3m$
$h = h_{2} - h_{1} = 19.6 m$ $h=h_2-h_1=19.6m$
to fall down:
$- h_{2} = - \frac{1}{2} \cdot 9.81 t_{2}^{2}$ $-h_2=-1/2*9.81t_2^2$
$t_{2} = 2.13 s$ $t_2=2.13s$
into $v_{f} - v_{i} = a t$ $v_f-v_i=at$ :
with $v_{i} = 0$ $v_i=0$
$v_{f} = 21 \frac{m}{s}$ $v_f=21m/s$
and
$T_{c o m p . t e} = t_{1} + t_{2} = 2.87 s$ $T_(comp.te)=t_1+t_2=2.87s$

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Question #6f91b

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