In the time taken by the projectile to reach from `A` to `B` is `t`. Then the distance `AB` is equal to.
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Time taken by the particle to reach from A to B is t . Then the distance AB is equal to

If time taken by the projectile to reach Q is T , than PQ =

A projectile is fired horizontally with velocity of 98 m/s from the top of a hill 490 m high. Find (a) the time taken by the projectile to reach the ground, (b) the distance of the point where the particle hits the ground from foot of the hill and (c) the velocity with which the projectile hits the ground. (g= 9.8 m//s^2) .

From the top of a hill 480 m high , a projectile is fired horizontally with a speed of 96 m//s . Calculate (i) the time taken by projectile to reach the ground . (ii) the distance of the target from the hill . (iii) the velocity with which the projectile hits the ground . Take g=10ms^(-2)

For ground to ground projectile, time take by a particle to ge point O to point C is T_(1) and during the same motion time taken by the particle to ge from A to B is T_(2) then height 'h' is :

(a) A particle moving with constant acceleration from A to B in a straight line AB has velocities u and v at A and B respectively. Find the velocity of the particle at the midpoint of AB . (b) If the time taken by the particle to go from A to the midpoint of AB is two times that from the midpoint of AB to B then find the value of v//u .

The time taken by a particle in reaching from trough to crest in a transverse wave is -

A stone is projected from point A with speed u making an angle 60^@ with horizontal as shown. The fixed inclined suface makes an angle 30^@ with horizontal. The stone lands at B after time t. Then the distance AB is equal to

The given figure shows the elliptical orbit of a planet m about the sun S. If t_(1) is the time taken by the planet of move from C to D and t_(2) is the time taken by the planet to move from A to B, then the relation between t_(1) and t_(2) is [Given, area SCD = 2 (area SAB)]