3371) v2=u2+2as
here, v= 1/4v and u= 1/3 v and s=28m a=g= -10m/s2 [-ve as it is retardation.]
Find v2.
Again v2=u2+2as
Now is the second case, when we throw up the ball. Here, v2=0 and u2=the value of the previous v2.
Find s.
Doing all the calculations, ans most probably is:
576 m
337
Sayan Sinha
·2013-05-09 10:15:01
2) For the first case,
v2=u2+2as
Here, v=0; a=10 m/s2; s=15 m.
Find u.
u comes out to be 17.32
The objects meet each other after some particular time. Now, for the ball, let distance be S1. Since total distance= 15m, therefore for the stone, the distance after which they will meet is 15-S1 m.
We know, S= ut+ 12at2
Replace with values.
For the ball, S= S1; u=17.32; a=10.
For the stone, S=15-S1; u=0; a=10.
So the equations are:
S1= 17.32t -5t2
15-S1= 5t2
Add the equations.
t comes out to be 0.87
Therefore S1=10.75
So, 15-S1 is= 4.25
So, their ratio= 10.754.25= 2.53...
337
Sayan Sinha
·2013-05-09 10:21:11
3) cos 45° = basehypotenuse
=> 1√2=base84
So, base= 59.4
Therefore, ans= 59.4 m every second...
337
Sayan Sinha
·2013-05-09 10:33:37
4) Hint: (20h1-10h2)2=10(2h1-h2)
Solve for h.
337
Sayan Sinha
·2013-05-09 10:34:56
4) Hint: (20h1-10h2)2=10(2h1-h2)
Solve for h.
