Question

A particle when thrown obliquely from ground , can have maximum horizontal range 'x'.When it is thrown vertically upwards with the same speed, the maximum height reached by the particle will be

Answer 1

Given :  A particle when thrown obliquely from ground , can have maximum horizontal range 'x' . it is thrown vertically upwards with the same speed,  

To find : maximum height reached by the particle

Solution:

maximum horizontal speed is achieved when  

object is thrown at angle of 45°

Let say  velocity magnitude was  V

Hence Vertical velocity = VSin45 = V/√2

& Horizontal Velocity = VCos45 = V/√2

Time to reach max height  ( at top Velocity = 0)

using V = U + at

=>  0  = V/√2  + (-g)t

=> t = V/g√2

Total time of flight = 2 * V/g√2  = V√2 / g

Horizontal Distance covered = (V/√2) (V√2 / g)

= V²/g

=> x = V²/g

Now object thrown Vertically with velocity V

using V² - U² = 2aS

=> 0 - V²  = 2(-g)H

=> H = V²/2g

=> H = (V²/g)/2

=> H = x/2

maximum height reached by the particle will be x/2

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