In a rescue attempt,a stationary helicopter in air drops a life preserver to a drowning man being swept downstream by a river current of constant velocity "v" .The helicopter is at a height of "45m" .The man is "6.0m" upstream from a point directly under the helicopter when the life preserver is released.It lands "3.0m" in front of the man.How fast is the current flowing? Neglect air resistance.`g=10m/s^(2)`

A child in danger of drowning in a river is being carried downstream by a current that flows uniformly at a speed of 2.5 km//h. The child is 0.6 km from shore and 0.8 km upstream of a boat landing when a rescue boat sets out. If the boat proceeds at its maximum speed of 20 km//h with respect to the water, what angle does the boat velocity v make with the shore? How long will it take boat to reach the child?

A helicopter is moving verticallly upwards with a velocity 5 ms^-1 . When the helicopter is at a height 10 m from ground, a stone is thrown with a velocity (3 hat i + 4 hat j) m s^-1 from the helicopter w.r.t. the man in it. Considering the point on ground vertically below the helicopter as the origin of coordinates, and the ground below as xy plane, find the coordinates of the point where the stone will fall, its distance from origin at the instant the stone strikes the ground, assuming helicopter moves upwards with constant velocity. .

A cracker is thrown into air with a velocity of 10 m//s at an angle of 45^(@) with the vertical. When it is at a height of 0.5 m from the ground, it explodes into a number of pieces which follow different parabolic paths. What is the velocity of centre of mass, when it is at a height of 1 m from the ground? ( g = 10 m//s^(2) )

A body is thrown with the velocity v_0 at an angle of theta to the horizon. Determine v_0 "in" m s^-1 if the maximum height attained by the body is 5 m and at the highest point of its trajectory the radius of curvature is r = 3 m . Neglect air resistance. [Use sqrt(80) as 9] .

A stone is projected from level ground at t=0 sec such that its horizontal and vertical components of initial velocity are 10(m)/(s) and 20(m)/(s) respectively. Then the instant of time at which tangential and normal components of acceleration of stone are same is: (neglect air resistance) g=10(m)/(s^2) .

A particle Is projected at point 'A' with initial velocity 5 m//s at an angle theta = 37^(@) with the vertical y axis. A strong horizontal wind gives the particle a constant horizontal acceleration 6 m//s^(2) in the x direction. If the particle strikes the ground at a point directly under its released position, determine the height 'h' of point A . The downward acceleration is taken as the constant g = 10 m//s^(2) (take sin37^(@) = (3)/(5), cos37^(@) = (4)/(5) )

(i) A block is connected to an ideal spring on a horizontal frictionless surface. The block is pulled a short distance and released. Plot the variation of kinetic energy of the block vs the spring potential energy. (ii) A ball of mass 200 g is projected from the top of a building 20 m high. The projection speed is 10 m/s at an angle theta=sin^(-1)(3/5) from the horizontal. Sketch a graph of kinetic energy of the ball against height measured from the ground. Indicate the values of kinetic energy at the top and bottom of the building and at the highest point of the trajectory, specifying the heights on the graph. Neglect air resistance and take g = 10 m//s^(2)