A projectile is thrown at angle of `30^(@)` with a velocity of `10m/s` .The change in velocity during the time interval in which it reaches the highest point is

A particle of mass m is projected at an angle alpha to the horizontal with an initial velocity u . The work done by gravity during the time it reaches its highest point is

A particle of mass 1 kg is projected at an angle of 30^(@) with horizontal with velocity v = 40 m/s . The change in linear momentum of the particle after time t = 1 s will be (g = 10 m// s^(2) )

A body is thrown vertically upwards with a velocity of 98ms^(-1) Calculate (i) the maximum height reached (ii) the time taken to reach the highest point (iii) the velocity at a heighest of 196m from the point of projection (iv) the velocity with which it returneds to the ground and (v) the time taken to reach the ground.

A projectile is thrown upward with a velocity v_(0) at an angle alpha to the horizontal. The change in velocity of the projectile when it strikes the same horizontal plane is

A projectile is thrown with a velocity u_(0) at an angle theta with the horizontal. The ratio of the rate of change of speed w.r.t. time, at highest point to that at the point of projection, is

A projectile is thrown with an initial velocity of (xveci + y vecj) m//s . If the range of the projectile is double the maximum height reached by it, then

A projectile of mass 100 g is fired with a velocity of 20 ms^(-1) making an angle of 30^@ with the borizontal. As it rises to the highest point of its path its momentum changes by .

A projectile is thrown with an initial velocity of vecv = (phati+qhatj) m/s. If the range of the projectile is four times the maximum height reached by it, then

A projectile is thrown with a velocity of 10sqrt(2)m//s at an angle of 45^(@) with horizontal. The interval between the moments when speed is sqrt(125)m//s is (g=10m//s^(2))