a projectile is thrown from origin in xy plane with a speed of 10 metre per second at an angle of 45 degree above positive x-axis where positive x-axis is horizontal Axis y axis vertical axis in origin is at ground assuming air resistance to be negligible and G is equal to 10 metre per second the minimum distance of projectile from point P having coordinate (5,5)
Answers
Answer:
5/2
Explanation:
a projectile is thrown from origin in xy plane with a speed of 10 metre per second at an angle of 45 degree
Horizontal Speed = 10Cos45 = 5√2 m/s
Vertical Speed = 10Sin45 = 5√2 m/s
Horizontal Distance covered in time t = 5√2 t m
Vertical distance covered = 5√2 t - (1/2)10t² = 5√2 t - 5t²
at any time t the coordinates of Projectile
(5√2 t , 5√2 t - 5t²)
Distance from (5 , 5)
= √ (5√2 t - 5)² + (5√2 t - 5t² - 5)²
(5√2 t - 5)² + (5√2 t - 5t² - 5)² should be minimum
differentiating wrt t
2 (5√2 t - 5)5√2 + 2(5√2 t - 5t² - 5)(5√2 - 10t) = 0
=> (√2 t - 1)√2 + (√2t - t² - 1)(√2 - 2t) = 0
=> (2t - √2) + (√2t - t² - 1)(√2 - 2t) = 0
=> (√2 - 2t) (√2t - t² - 1 - 1) = 0
=> t = 1/√2
or t² - √2t + 2 = 0
=> t is imaginary
putting t = 1/√2
Distance = √ (5 - 5)² + (5 - 5/2 - 5)² = 5/2
minimum distance of projectile from point P having coordinate (5,5) = 5/2