Question

Three balls are projected with same speed . ball A is thrown vertically downward, ball B is thrown vertically upward and Ball C is thrown horizontally ta, tb and tc are respective time taken by the ball A, B and C to reach ground find the correct answer from the following choices


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Answer 1

Answer:

Explanation:opt b is ans

as u see tc is horizontal so it is proportional to ta and tb as thes are up and down dirn but it is not directly proportional so sq root is there

Answer 2

Given that,

Ball A is thrown vertically downward.

Ball A is thrown vertically upward.

Ball A is thrown horizontally.  

We need to calculate the distance

Using equation of motion

For ball A,

The distance is

$s=ut_{a}-\dfrac{1}{2}gt_{a}^2$.....(I)

For ball B,

The distance is

$s=ut_{b}+\dfrac{1}{2}gt_{b}^2$....(II)

For ball C,

The distance is

$s=ut_{c}+\dfrac{1}{2}gt_{c}^2$

Ball C is thrown horizontally so the initial velocity will be zero

$s=\dfrac{1}{2}gt_{c}^2$....(III)

Now, Put the value of s in equation (II)

$\dfrac{1}{2}gt_{c}^2=ut_{b}+\dfrac{1}{2}gt_{a}^2$

$\dfrac{1}{2}g(t_{c}^2-t_{b}^2)=ut_{b}$

$t_{c}^2-t_{b}^2=\dfrac{2ut_{b}}{g}$.....(IV)

Now, Put the value of s in equation (I)

$\dfrac{1}{2}gt_{c}^2=ut_{a}-\dfrac{1}{2}gt_{a}^2$

$\dfrac{1}{2}g(t_{c}^2+t_{a}^2)=ut_{a}$

$t_{c}^2+t_{a}^2=\dfrac{2ut_{a}}{g}$.....(V)

Divided equation (IV) by equation (V)

$\dfrac{t_{c}^2-t_{b}^2}{t_{c}^2+t_{a}^2}=\dfrac{\dfrac{2ut_{b}}{g}}{\dfrac{2ut_{a}}{g}}$

$\dfrac{t_{c}^2-t_{b}^2}{t_{c}^2+t_{a}^2}=\dfrac{t_{b}}{t_{a}}$

$t_{a}t_{c}^2-t_{a}t_{b}^2=t_{b}t_{c}^2+t_{b}t_{a}^2$

$t_{a}t_{c}^2-t_{a}t_{b}^2-t_{b}t_{c}^2-t_{b}t_{a}^2=0$

$t_{c}^2(t_{a}-t_{b})-t_{a}t_{b}(t_{b}+t_{a})=0$

$t_{c}^2=\dfrac{t_{a}t_{b}(t_{b}+t_{a})}{(t_{a}-t_{b})}$

$t_{c}^2= t_{a}t_{b}$

$t_{c}=\sqrt{t_{a}t_{b}}$

Hence, The correct answer is $t_{c}=\sqrt{t_{a}t_{b}}$

(B) is correct option.