A body is projected up with a velocity u.It reaches a point 'p' its path at times t_(1) and t_(2) seconds from the time of projection.Then (t_(1)+t_(2))

A body is projected vertically upwards with some velocity u . It at the same point at t = 6 s as was at t = 4 s . Find the value of u .

A body is projected upwards with a velocity u . It passes through a certain point above the grond after t_(1) , Find the time after which the body posses thoruth the same point during the journey.

A stone projected vertically up from the ground reaches a height y in its path at t_(1) seconds and after further t_(2) seconds reaches the ground. The height y is equal to

A body is projected vertically upwards at time "t=0" and it is seen at a height H at instants t_(1) and t_(2)seconds during its flight.The maximum height attained is (g is acceleration due to gravity

A body projected vertically upward from the top of a tower reaches the ground in itme t_(1). If it is projected vertically downwards from the same position with the same velocity , it reaches the ground in time t_(2) . If a body is released from rest and from the same position then what will be the time (t) required by body to reach the ground?

Two bodies are projected from the same point with same speed in the directions making an angle alpha_(1) and alpha_(2) with the horizontal and strike at same point in the horizontal plane through a point of projection. If t_(1) and t_(2) are their time of flights.Then (t_(1)^(2)-t_(2)^(2))/(t_(1)^(2)+t_(2)^(2))

For a given velocity, a projectile has the same range R for two angles of projection. If t_(1) and t_(2) are the time of flight in the two cases, then t_(1) = t_(2) is equal to

A body is thrown vertically up to reach its maximum height in t seconds. The total time from the time of projection to reach a point at half of its maximum height while returning (in seconds) is