Question

a ball A is projected from the ground such that is horizontal range is maximum.another ball B is dropped from height equal to the maximum range of A.the ratio of the time of flight of A to the time of fall of B is

Answer 1

the answer is 1:1

explanation: if ranges are equal ,time are also equal

Answer 2

Answer:

1:1

Explanation:

We are given that

Ball A is projected from the ground and its horizontal range is maximum.

Ball B is dropped from height =Maximum range of ball A.

We have to find the ratio of the time of flight of A  to the time of fall of B.

For ball A

T=$\frac{2 usin\theta}{g}$

When the horizontal range is maximum then $\theta=45^{\circ}$

T=$\frac{2 usin45}{g}=\frac{2u}{\sqrt2g}=\frac{1.41u}{g}$

$sin 45^{\circ}=\frac{1}{\sqrt2}$

Maximum range=$\frac{u^2}{g}$

For ball B

u=0, $s=\frac{u^2}{g}$

$S=ut+\frac{1}{2}gt^2$

$\frac{u^2}{g}=0+\frac{1}{2}gt^2$

$t^2=\frac{u^2}{g}\times \frac{2}{g}=\frac{2u^2}{g^2}$

$t=\sqrt{\frac{2u^2}{g^2}}=\frac{1.41u}{g}$

$\frac{T}{t}=\frac{\frac{1.41u}{g}}{\frac{1.41u}{g}}$

$\frac{T}{t}=1/1$

T:t=1:1

Hence, the ratio  of time of flight of A to the time of fall of B=1:1