Question

a pebble is thrown horizontally from the top of 20 m high tower with initial velocity of 10 m in the air drag is a negligible the speed of the people when it is at the same distance from the top as well as base of the tower

Answer 1

answer will be 2u^2/g * (tan alpha)/(cos alpha)

Answer 2

Given is the data of the height of the tower h and the initial velocity u as 20 m and $10\frac { m }{ s }$ respectively. We have been asked to find out the speed at the equidistance from top of the tower and base of the tower which will be $\frac { 20 }{ 2 } =10m$. By applying equation of motion, we get – $v^2 = u^2 + 2as$, since the horizontal velocity will not changed.

We just need to find the vertical component of the thrown pebble, which is when the initial velocity will be zero vertically, therefore $v^2 = 2as$ where a is acceleration due to gravity and s is the displacement. $v_{ y }=\sqrt { 2 } \times 10\times 10=100\sqrt { 2 } \frac { m }{ s }$. Therefore the speed of the thrown pebble be – $speed =\sqrt { (v_{ x }^{ 2 }+v_{ y }^{ 2 }) } \Rightarrow speed=\sqrt { (100+200) } =10\sqrt { 3 } \frac { m }{ s } .$