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Trajectory Of Horizontal Projectile...

Trajectory Of Horizontal Projectile

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Velocity at a general point P(x, y) for a horizontal projectile motion is given by v=sqrt((v_("x")^(2)+v_("y")"^(2))),tan alpha=(v_(y))/(v_(x)) alpha is angle made by v with horizontal in clockwise direction Trajectory equation for a horizontal projectile motion is given by x=v_(x)t=ut y=- (1//2)"gt"^(2) eliminating t, we get y=-(1//2)(gx^(2))/(u^(2) An aeroplane is flying horizontally with a velocity of 720 km/h at an altitude of 490 m. When it is just vertically above the target a bomb is dropped from it. How far horizontally it missed the target?

elocity at a general point P(x, y) for a horizontal projectile motion is given by v=sqrt((v_("x")^(2)+v_("y")"^(2))),tan alpha=(v_(y))/(v_(x)) alpha is angle made by v with horizontal in clockwise direction Trajectory equation for a horizontal projectile motion is given by x=v_(x)t=ut y=- (1//2)"gt"^(2) eliminating t, we get y=-(1//2)(gx^(2))/(u^(2) 5An aeroplane is in a level flying at an speed of 144 km//hr at an altitude of 1000 m. How far horizontally from a given target should a bomb be released from it to hit the target ?

The trajectory of a projectile fired horizontally with velocity u is parabola given by-

The trajectory of a projectile in a vertical plane is y = ax - bx^2 , where a and b are constant and x and y are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.