Let the angle of projection are θ and 90°-θ. Let u be the initial velocity of cannon ball.∴ Horizontal range, R=u2gsin2θ=2u2gsinθcosθ .....1For angle of projection θ, height, h1=u2sin2θ2g .....2For angle of projection 90°-θ, h2=u2sin290°-θ2g=u2cos2θ2g .....3h1h2=u4sin2θcos2θ4g2 From equation 2 and 3⇒h1h2=u2sinθcosθ2g=14R From equation 1⇒R=4h1h2
Tardigrade - CET NEET JEE Exam App
© 2022 Tardigrade®. All rights reserved |