A body is thrown vertically upwards from the top of the tower. It reaches the ground in t1 seconds. If it is thrown vertically downwards from the same point with the same speed, it reaches the ground in t2 sec. Please prove If it is allowed to fall freely from the top of the tower, then the time it takes to reach the ground t is given by formula t=√(t1*t2).

Solution.

Formula. Yf=Yi+Vi*t+a*t^2/2; Yf – final position; Yi – initial position; a=-g=-9.81 m/s^2; Yi=h – height of the tower; Vi – initial velocity

The body thrown vertically upwards .

0=h+v*t1-g*t1^2/2

The body thrown vertically downwards .

0=h-v*t2-g*t2^2/2

The body fall freely.

0=h-g*t^2/2 ; So, h=g*t^2/2.

Let’s use it for the first and second equations.

0=g*t^2/2+v*t1-g*t1^2/2

0=g*t^2/2+v*t2-g*t2^2/2

From the second equation

v=(g*t^2-g*t2^2)/(2*t2).

Use it for the first equation.

0=g*t^2/2+(g*t^2-g*t2^2)/(2*t2)*t1–g*t1^2/2 ; Divide by g

0=t^2/2+(t^2-t2^2)/(2*t2)*t1-t1^2/2; Multiply by 2

0=t^2+(t^2-t2^2)/t2*t1-t1^2; Multiply by t2

0=t^2*t2+t^2*t1-t2^2*t1-t1^2*t2

0=t^2*(t1+t2)-t1*t2*(t1+t2) ; Divide by (t1+t2)

0=t^2-t1*t2

And Finally

t=√(t1*t2)

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