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# Kinematics. Throwing body up problem

Kinematics problem from the Svetozarov's book for MEPI matriculates

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Well, we finally got to the acceleration topic, which, as you know, is the change in speed over time.

Condition:
The second body was thrown vertically upward next to the first one after t with the same speed v. How long after throwing the second body and at which height h the two bodies will collide?

Solution:
Distance traveled equation
$s(t)=v_0t+\frac{at^2}{2}$
For the first body
$y(t)=v(t+t_1)-\frac{g(t+t_1)^2}{2}$
For the second body
$y(t)=vt_1-\frac{gt_1^2}{2}$
Accordingly, when they meet, their coordinates will coincide, i.e.
$v(t+t_1)-\frac{g(t+t_1)^2}{2}=vt_1-\frac{gt_1^2}{2}$,
from which
$t_1=\frac{v}{g}-\frac{t}{2}$
After finding time substitute it in any formula for the distance and find h
Gravitational acceleration is assumed equal to 9.8 m/s2

### Kinematics. Throwing body up problem

Digits after the decimal point: 2
Time to collision t1, (s)

Collision height h, (m)