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# Kinematics. Throwing body up problem

Kinematics problem from the Svetozarov's book for MEPI matriculates

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)

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