Question

a ball thrown horizontally with velocity v from the top of a tower of height h reaches the ground in t second.If another ball of double the mass is thrown horizontally with velocity 3v from the top of another tower of height 4h it reaches the ground in(seconds) if first ball reaches the the ground at a horizontal distance d then the second ball reaches the ground at horizontal distance

Answer 1

Answer: the distance of the second ball is 6d

Explanation: please refer to the text material.

Answer 2

Answer:

Distance of second ball, D = 6d

Explanation:

In the question,

We know that,

For first ball :  

Horizontal velocity of the ball = v

Time taken = t seconds

Height = h

So,

Time taken to reach the bottom is given by,

Using Second equation of motion

$s=ut+\frac{1}{2}at^{2}$

So,

$h=(0)t+\frac{1}{2}gt^{2}\\So,\\t=\sqrt{\frac{2h}{g} }$

Now,

Distance covered in horizontal, d is given by,

$d=(v)t+\frac{1}{2}(0)t^{2}\\d=vt$

So,

On putting the value of 't' in the equation we get,

$d=v\sqrt{\frac{2h}{g}}$

Now,

For the second ball :

Horizontal velocity = 3v

Height = 4h

Time taken = T

Distance from ground = D

Using Second equation of motion

$T=\sqrt{\frac{2\times 4h}{g}} =2\sqrt{\frac{2h}{g}}$

Now,

Distance travelled in horizontal,

$D=(3v)(T) +\frac{1}{2}(0)T^{2}\\D=3vT$

On putting the value of T we get,

$D=3v\times 2\sqrt{\frac{2h}{g}}\\D=6v\sqrt{\frac{2h}{g}}$

So, on comparing both the distances we can say that,

$\frac{d}{D}=\frac{v\sqrt{\frac{2h}{g}}}{6v\sqrt{\frac{2h}{g}}}\\ So,\\\frac{d}{D}=\frac{1}{6}\\D=6d$