A stone is thrown in a vertically upward direction with a velocity of 5m s^-1. If acceleration of the stone in the downward direction 10m s^-2, what will be the height attained by the stone and how much time will it take to reach there?

Question

anonymous · Answer

$$V_{int} = 5 ms^{-1}$$
$$a_g = -10 ms^{-2} $$
At the highest point $$V_{final} = 0$$
$$V_{final} = V_{int} + a_gt$$

anonymous · Answer

or Use this one, this one is much better $$V_{final}^{2} - V_{int}^2 = 2a_g H$$

anonymous · Answer

sorry took wrong values 

$$ 25 = 20 \times H$$
$$H = \frac{25}{20}$$

anonymous · Answer

this is the correct one

anonymous · Answer

did you get that ?

anonymous · Answer

no

anonymous · Answer

i think i got the correct answer
is it s= 1.25 m
and t= 0.5 sec?

anonymous · Answer

it is right ! hit the good answer

anonymous · Answer

First the time.  $$V _{o}+a \times t =0$$
So 5m/s-10m/s^2(t)=0
5m/s=10m/s^2(t)
.5s=(t)
$$D =(V _{o}+V)/2\times t$$
D=(5m/s+0m/s)/2*.5=5m/s/4=1.25m

anonymous · Answer

ok.....
u=5m/s
a=10 m\(s^2)
s=ut+(1/2)a(t^2)
and 
t=usin(t)/g
now t=90degree
so
t=5/10= 0.5s
and 
s=5*0.5+(1/2)(10*(0.5^2)= 2.5+(0.5*0.25*10)=2.5+1.25= 3.75m

anonymous · Answer

ohh sorry
since at height, u=0... so. s=(1/2)a(t^2)=(1/2)*10*(0.5^2)=1.25m