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
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Answers
Step-by-step explanation:
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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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