Question

In hostel playground, Jiyang hits a ball directly aiming the upper parts of a square window (BCDE) at angle
45 with initial velocity u =10m/sec. The ball crosses the window horizontally at its lower parts. The area of
window is​

Answer 1

Given :  In hostel playground, Jiyang hits a ball directly aiming the upper parts of a square window (BCDE) at angle  45 with initial velocity u =10m/sec. The ball crosses the window horizontally at its lower parts.  

To find :  Area of window

Solution:

Let say mid point of EF is  M  which is the maximum height of the Ball

Vertical velocity =  10 Sin45°   = 10/√2 = 5√2  m.s

Time taken to reach at peak =  t

V = U + at

a = g

=> 0  = 5√2 + (-10)t

=> t =  1/√2

Vertical distance  S   using V² - U² = 2aS

=> 0² - (5√2)²  = 2(-10)S

=> S = 2.5   m

Height of  point window from ground = 2.5 m where

Horizontal Velocity =10 cos45 = 10/√2 = 5√2  m.s

Horizontal Distance  =  5√2 *( 1/√2)  =  5  m

Jiyang hits a ball directly aiming the upper parts of a square window (BCDE)

Hence  tan 45°   =  Height of top of window from ground / 5

=> Height of top of window from ground = 5 m

Height of bottom from ground  = 2.5 ,  

Height of Window  =  5 - 2.5 = 2.5  m

Area of Square window = 2.5 * 2.5  = 6.25  m²

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