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A projectile is projected from ground, such that 2 second before it reaches the highest point, it makes an angle 37 degrees with vertical, then the velocity of projectile at highest point is (1 ) 5 m/s (2) 10 m/s (3) 15 m/s ( 4) 20 m/s |
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Answer» Velocity 2 seconds before it reaches maximum height=velocity 2 seconds after it reaches maximum height Vy= uy-gt = -2g = 20 m/s The angle made by the vertical is 37 but to find tan we need angle with the horizontal so the needed angle is 90 - 37 = 53. So, tan 53=Vy/Vx 4/3=20/Vx Vx=20×3/4 =15 m/s |
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