Question

From the top of a tower a ball is projected horizontally with a velocity u. If the magnitudes of the horizontal and vertical displacements of the ball are to be equal during the motion of the ball, what should be the minimum height of the tower (g is acceleration due to gravity)

Answer 1

Answer: h = u^2 / 2g

Step-by-step explanation:

Given the following :

Horizontal Velocity of projection= u

If :

magnitude of horizontal = magnitude of vertical Displacement

u = u

Minimum height of tower (h) =?

Horizontal Velocity = u

Gravity potential energy = mgh - - - (1)

Kinetic energy = 1/2 mu^2 - - - (2)

m = mass

Where u = magnitude of velocity

g = acceleration due to gravity

h = height

Equating (1) and (2)

mgh = 1/2 mu^2

gh = 1/2 mu^2

2gh = u^2

u = √2gh

Vertical component of Velocity 'v' will be:

u = √2gh

u = √2 × g × h

Square both sides

u^2 = 2 × g × h

h = u^2 / 2g