Question

A body is thrown horizontally with a velocity √2gh from the top of a tower of height h. It strikes the level ground through the foot of the tower at a distance x from the tower. The value of x is : (1) h (2) h/2 (3) 2h (4) 2h/3 plz help !​

Answer 1

Answer:

2h

Explanation:

---velocity(√2gh with a)--->

h

<-----------------x---------------->

           Ground

Time taken to reach the end of x is same in both the cases(either horizontally or vertically).

       Vertically:

there is no velocity, u = 0

S = height of tower = h

time taken = t

      ∴ S = ut + 1/2 at²

         h = (0)t + 1/2 gt²

        √(2h/g) = t      

          Horizontally:

There is no acceleration, a = 0

u = (√2gh),    S = x,   t = √(2h/g)

           ∴ S = ut + 1/2 at²

⇒ x = √(2gh)*√(2h/g) + 1/2 (0)t²

⇒ x = √(2gh) * √(2h/g)

⇒ x = 2h

     Therefore the required value of x is 2h.

Answer 2

Explanation:

Given Horizontal velocity = VH = √2gh,

Vertical velocity=Vv = 0

Using S = ut + 1/2at²

⇒− h = 0 - 1/2gt²

t⇒√2h/g

Therefore, x = Vh × t

⇒x = √2gh × √2h/g = 2h