Question

A bomb is fired horizontally with a velocity of 20m/s from the top of the tower 40m high.After how much time and at what horizontal distance from the tower will the bomb strike the ground.

Answer 1

Assuming it to be projectile motion(It is) :
u=20m/s
h=40m
Therefore time of flight T=
$\sqrt{2h \div g}$
Hence T=
$\sqrt{8}$
Now R(Range) =
$u \times t = 20 \sqrt[]{8}$

Answer 2

Let us assume, horizontal distance is x.

We know height of tower (h) = 40 m

                                            u = 0

                                $g = 9.8\times \frac { m }{ { s }^{ 2 } }$

                                $h=ut+\frac { 1 }{ 2 } g{ t }^{ 2 }$

                                $40=0+\left( \frac { 1 }{ 2 }  \right) \times 9.8\times { t }^{ 2 }$

                                ${ t }^{ 2 }=40\times \frac { 2 }{ 9.8 }$

                                t = 2.86

Given,

                                $v=20\times \frac { m }{ s }$

                                $Distance = Speed \times Time$

                                =$20 \times 2.8=57.2m$