The motion of a stone thrown into a pond is described by where t is time express
ID: 1816135 • Letter: T
Question
The motion of a stone thrown into a pond is described by where t is time expressed in s, and t = 0s is the time whenthe stone first hits the water. Determine the stones velocity andacceleration. In addition, find the initial angle of impact of the stone with the water, i.e., the angle formed betweeen thestone's trajectory and the x axis at t = 0s. The x axis being thesurface of the water. I was able to find the velocity and acceleration bydifferentiation, but I can't figure out how to get the angle ofimpact . The motion of a stone thrown into a pond is described by R(t)=[(1.5-0.3e^-13.6t)i+(0.094e^-13.6t-0.094-0.72t)j]m where t is time expressed in s, and t = 0s is the time when the stone first hits the water. Determine the stones velocity and acceleration. In addition, find the initial angle of impact theta of the stone with the water, i.e., the angle formed betweeen the stone's trajectory and the x axis at t = 0s. The x axis being the surface of the water. I was able to find the velocity and acceleration by differentiation, but I can't figure out how to get the angle of impact theta .Explanation / Answer
Step 1: -------------- First of all, we derive to get the velocities. v(t) = dr(t)/dt = 4.08e-13.6 t i +(-1.2784e-13.6t -0.72)j Step 2: --------------- If we divide v0y / v0x we get v0y v0 sin ----- = -------------- = tan v0x v0cos Step 3: Preparing components of initial velocity ----------------------- If t=0, v0x = vx(0) = 4.08 e -13.6 (0) = 4.08 m/s v0y = vy (0) = -1.2784e-13.6(0) -0.72 = -1.9984 m/s Step 4: Returning to step 2: ----------------------------- Remembering tan = v0y / v0x we have tan = -1.9984 /4.08 and = 360° - tan-1 (1.9984/4.08) =333.9° Another way to put the answer is 26.1° with the horizontal inthe fourth quadrant.