A 0.80-kg block is held in place against the spring by a 67-N horizontal external force (see the figure). The external force is removed, and the block is projected with a velocity v1 = 1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity v2 = 1.9 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction over this section is 0.39. The velocity of the block is v3 = 1.4 m/s at C. The block moves on to D, where it stops. The spring constant of the spring is closest to

A., d., and f i think

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.