Question

A ball is dropped from a height h above ground.
Neglect the air resistance, its velocity (v) varies
with its height y above the ground as
(1) 2g(h-y) (2) 2gh
(3) Vagy
(4) 2g(n+y)​

Answer 1

Answer:

Explanation:

Given A ball is dropped from a height h above ground.

Neglect the air resistance, its velocity (v) varies

with its height y above the ground as

The ball travelled a distance h-y when the ball is above the ground.Now we have the equation of accelerated motion we get

V^2 = 2g(h – y)

2gy = 2gh – v^2

y = 2gh – v^2/2g -----------1

So this relation is between v and y.  

Answer 2

Answer:

√2g(h-y)

Explanation:

A ball is dropped from a height h above ground.

Neglect the air resistance, its velocity (v) varies

with its height y above the ground as

(1) 2g(h-y) (2) 2gh

(3) Vagy

(4) 2g(n+y)​

Using The Formula

V² - U² = 2aS

U = intial Velocity = 0 (as deopped)

V = Velocity when Distance covered = S

a = g

V² = 2gS

height y above the ground = Height - Distance Covered

=>  y = h - S

=> S = h - y

using this Value

V² = 2g(h - y)

V = √2g(h - y)