has An apparatus involving a spring and projectile is set up

has An apparatus involving a spring and projectile is set up according to the below. The spring an unstretched length of 12.0 cm. The total initial into account both potential energy ne to take elastic and gravitational potential energy) How fast is the projectie moving at arbitrary point, P? How fast is the projectile moving just before striking the ground at point, Spring Constant 4.0 cm Equals 98.0 Nm

Solution

The potential energy of the spring is = k x^2 / 2 , x is the length compressed length, given total length = 12cm and 4cm after compression , so compressed length is 8 cm = x

through of the journey of the projectile the horizantal velocity will not change , and the projectile will gain velocity with time it fall to ground.

the speed of the projectile at point p = sqrt (Vx^2 + Vy^2) with Vx, Vy   velocitys in horizantal and vertical directions

we have to determine Vx and Vy tto solve speed at p

from the spring the projectile gains Vx from its potential energy   kx^2/2 = mVx^2 / 2

                                               k spring constatn x is compressed length , m is mass of projectile , Vx is velocityu gained

                       Vx = sqrt ( kx^2 / m) = 2.639 m/ sec

            Vertical velocity of the ball after falling a height h, let be Vy , such that mgh = mVy^2 / 2

                                               Vy = sqrt (2gh ) = 2.257 m /sec

        so the speed at point P is = sqrt (Vx^2 + Vy^2) = 3.472 m/sec

teh velocity of the ball just before striking the ground V = (Vx^2 + Vy^2) with change in Vy = sqrt (gh) and h = 46 cm

                                                        Vx remains the same   and Vy = 3.002 m/se

                    velocity before hitting ground V = sqrt (2.639^2 + 3.002 ^2 ) = 3.997 m/sec