PROJECTILE MOTION
Question: If a water particle (droplet) in a stream of water in a fountain take 0.35s to travel from spout to receptacle when shot at an angle of 67 degrees at an initial speed of 5.0m/s, what is the vertical distance between the levels of the fountain?
y=1/2(g)t^2+Voyt
y=1/2(-9.8m/s^2)(0.35s)^2+4.6(0.35s)
(5.0m/s)sin67
y=1.0m
y=1/2(g)t^2+Voyt
y=1/2(-9.8m/s^2)(0.35s)^2+4.6(0.35s)
(5.0m/s)sin67
y=1.0m
Graph Explanation: Velocity
Explanation of Vyt graph: the initial velocity of the droplet is 5.0m/s. This is a Vy graph-vertical- 5.0sin67 making the initial velocity 4.6m/s. The droplets travels at 4.6m/s (inital vel.) from the spout to the receptacle in 0.35s. The velocity decreases with time because when the droplet reaches the receptacle, the velocity is 0m/s. (The slope is negative because the velocity decreases with time-the particle starts at 4.6m/s at the spout and ends at 0m/s in the receptacle.
Graph Explanation: Position
Explanation: the water particle (droplet) gains distance as time passes and it stops at 1.0m after 0.35s pass. The graph ends flat because the final velocity is 0m/s at .35s. The graph is curved because the droplet slowly comes to a stop. It starts out gaining distance at a slower rate at time goes on -and stops gaining distance at 0.35s.
How Does the Equation Best Describe the Motion
The equation y=1/2(g)t^2+Voyt best describes the motion because the distance (y) is determined by how long the object stays in the air 1/2(g)t^2 plus the initial vertical component of velocity x time (Voyt). The motion is accelerated by gravity (-9.8m/s^2 refers to the acceleration due to earth's gravity field). The initial velocity was 5.0m/s making the inital vertical component (Voy) 4.6 [5.0sin67] (sin was used because I had to find the vertical component) 67 degrees being the angle at which the droplet was shot. The time given is 0.35s which is how long it takes for the droplet to travel from the spout to the receptable. These values plugged into this formula determines the vertical distance between the levels of the fountain which is 1.0m.
1.0m is the distance the droplet covers in 0.35s travelling at the initial
vertical velocity of 4.6m/s and stopping in the receptable at 0m/s.
vertical velocity of 4.6m/s and stopping in the receptable at 0m/s.