I- a circus star of mass 75.0 kg is to launched from
groundlevel by a spring loaded device so that she clears a wall
3.00 mhigh and 4.00m away and lands onthe ground 5.00m behind
thewall.
what is the minimum impulse the spring must deliver to
achievethe desired trajectory?

The minimum impulse the spring must give the circus star in order to clear the wall height and land on the ground again is 287.25Ns. The spring needs to provide an impulse sufficient at leather get the circus performer to the highest point in flight after which she is now in free fall under the control of gravity.

Explanation:

The step by step solution for this problem can be found in the attachment below. Thank you.

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s