Tossing A Ball

The formula for vertical motion up is $v_{t} = v_{0}-gt$

Since the $v_{t}$ is 0 , so the formula becomes $\boxed{v_{0}=gt}$

$v_{1}=gt_{1}$

Then..

$v_{2}=gt_{2}$

$v_{2}=2gt_{1}$

So..

$\frac{v_{2}}{v_{1}}=\frac{2gt_{1}}{gt_{1}}$

$\frac{v_{2}}{v_{1}}=\frac{2}{1}= \boxed{2}$

We have $x=-\frac{g}{2}t^2+v_0t$ (Assuming the ball is thrown from $x=0$ ) . When the balls gets back to the ground: $x=0$ (with $t \neq 0$ ) then, $v_1=-\frac{g}{2}t_1$ and $v_2=-\frac{g}{2}t_2=-gt_1$ .
$\dfrac{v_2}{v_1} = \dfrac{-gt_1}{-\dfrac{g}{2}t_1}=2$ .

$v=u+at$

Given that acceleration due to gravity is the same for the two throws, we know that $v\propto t$ . This means that if you double $t$ , you also double $v$ .

Since ${ t }_{ 2 }=2\times { t }_{ 1 }$ ,

${ v }_{ 2 }=2\times { v }_{ 1 }$

$\therefore \frac { { v }_{ 2 } }{ { v }_{ 1 } } =2$

V=u-gt where u=0unit and g is constant !!

a formula for this case is:

t = s/v

t = time s = distance v = velocity

from above formula, we can take a conclude that if t is become twice bigger than before, then v is become a half of before.

for 1st ball 0=V1-gt1 V1=gt1

for 2nd ball 0=V2-gt2 V2=gt2

V2/V1=t2/t1=2

s=ut+at^{2}/2 s=o we get v=gt/2 hence v2 /v1=t2/t1 =2

If time taken for second ball is half as time taken for first ball then the speed of second ball should be twice as first ball ,

So speed of first ball be anything but it is half as second ball ,

So let the speed of first ball be 1(v1) then the speed for second ball equals to 2 (v2) ,

2(v2) / 1 (v1) = 2..

final velocity(v)=0 unit, v=v(1)-g t(1)...................(1) v=v(2)-g t(2)...................(2) 2/1 then,answer will be 2

when the ball moves up with velocity vi and when it is thrown 2nd time it moves up with velocity v2.the first ball takes time t and 2nd ball takes time t2 which is double so it means its velocity is also double .when we take any number instead of velocity its answer is always 2 like this v2/v1 when we take v2=10 and v1=5 because v2 is double that of v1 then 10/5 becomes equal to 2

for 1st ball 0=v1-gt1 v1=gt1

for 2nd ball 0=v2-gt2 v2=gt2

v2/v1=t2/t1=2

velocity 1st case is (at) in 2nd case is (2at) dividing 2nd by 1st =2.

as t2=2t1 => v2/v1=2t1/t1 => v2/v1=2

u can solve it through......v=u-gt

v2/v1 = t2/t1 ---> v2/v1 = 2t1/t1 = 2 [m/s]. Answer : 2

as we know that this comes under time of ascent so V=-GT V1=-9.81 T(T=T1) V2=-9.81 2T(T2=2T1) V2/V1=-9.81 T/-9.81 2T V2/V1=2

Well,the time and speed are directly proportionals,so if the T2 = 2xT1,we have the conclusion that V2 = 2xV1, so , 2/1 = 2.

for both cases final velocity ie V(f) will be equal to zero for the case 1 V(f)=v1 -gt 0=v1-gt for the case 2 v(f)=v2-g t1 0=v2-2g t1 v1/v2=2

v=u+at. In this case u is 0, thus, v=at thus v is directly proportinal to time. so,T2=2T1. So v2=2V1 v2/v1=2