Two Bodies are dropped from two different heights h1 and h2 one of them is dropped directly down while the other is projected with a horizontal velocity of 10m?s It is found that both reach the ground with same speed (g=10m/s^2) Then

h1/h2 = root10 h1-h2=5m

Question

Two Bodies are dropped from two different heights h1 and h2 one of them is dropped directly down while the other is projected with a horizontal velocity of 10m?s It is found that both reach the ground with same speed (g=10m/s^2)  Then

h1/h2 = root10                h1-h2=5m

anonymous · Answer

FOR BODY DROPPED DIRECTLY.
2ah1 = v^2 => v = sqrt(2gh1)----- = speed----2
For BODY 2.
Vertical velocity is sqrt ( 2gh2)
Horizontal is 10.
Therfore speed = sqrt ( 2*10*h2 + 100) = sqrt(20h2 + 100) ----1
 Equating 1 and 2,
20h1 = 20h2 + 100
h1 - h2 = 5

anonymous · Answer

@siddhantsharan  i did nt understand 2 nd eq

anonymous · Answer

Therfore speed = sqrt ( 2*10*h2 + 100) = sqrt(20h2 + 100)???????????

anonymous · Answer

The vertical component will be $$\sqrt{2*g*h2}$$

Horizontal is 10m/s.

Now these are two vectors. Therefore Their net resultant is 
$$\sqrt{a^2 + b^2}$$
Where a and b are the vertical and horizontal velocities,

anonymous · Answer

@siddhantsharan  ok  then how  eq 1  =  eq2

anonymous · Answer

Its given that the speeds are equal.

anonymous · Answer

ok sorry  i did nt see it thanzzzzz a lot

anonymous · Answer

@siddhantsharan   Vertical velocity is sqrt ( 2gh2)  how???

anonymous · Answer

$$v^2 - u^2 = 2as$$
You know that.
Now, s = h2 and u = 0
Therefore, $$v^2 = 2hg$$