Question

Q1. A stone thrown upward with a speed u from the top
of the tower reaches the ground with a velocity 5u. The
height of the tower is​

Answer 1

QUESTION-:

A stone thrown upward with a speed u from the top  of the tower reaches the ground with a velocity 5u. The  height of the tower is​

EXPLANATION-:

• Acceleration=g
• Final velocity(v)=5u
• Initial velocity(u)=u
• Height=?

Using the second equation of kinematics we have-:

$\leadsto \underline{\boxed{\bigstar \bf\; v^2=u^2+2gH}}$

Putting values in the formula we get-:

$\rightarrow \bf\; (5u)^2=u^2+2gH\\\\\rightarrow \bf\; 25u^2=u^2+2gH\\\\\rightarrow \bf\; 25u^2-u^2=2gH\\\\\rightarrow \bf\; 24u^2=2gH\\\\\rightarrow \bf\; \frac{24u^2}{2g} =H\\\\\dashrightarrow \boxed{\underline{\tt\dag \;\;\frac{12u^2}{g}=H}}$

Therefore the height of tower is -:

12u²/g