Question

A projectile is launched from level ground at an angle of 30 degrees to the horizontal. If the magnitude of the launch velocity is 30 m/s, calculate the time rate of change in speed and radius of curvature when t=1, 2, and when the projectile is at its max height.

Please do the problem how the description says to do it below and put the answer neatly in a table format. Thank you.

2) A projectile is launched from level ground at an angle of 30° to the horizontal.f the magnitude of the launch velocity is 30 m/s, calculate the time rate of change in speed (v) and the radius of curvature (p) when: a) t 1 seconds b) The projectile reaches its maximum height c) 2 seconds For each part, start with a sketch of the particle, the velocity and acceleration vectors, and the x-y coordinates. Next, overlay the n-t coordinates (Not Shown) with en and êt labels. This will help you see angles and how to project vectors to the n-t directions. ia) (example of how to start each part)

Answer 1

(a) at t = 1 v_x = 30 cos 30 degree = 25.98 m/s v_y = u_y + a_yt = 30 sin 30 degree - 10 times 1 v_y = 5 m/s theta prime = tan^-1 (5/25.98) = 70.89 degree now rate of change in speed = -g sin theta prime = 10 sin 10.89 degree = -1.89 m/s^2 radius of curvature rho = v^2/g cos theta prime = (25.98)^2 + 5^2/10 times cos 10.89 degree rho = 71.28 m (b) at highest point rate of change in speed v = 0 rho = v_x^2/g = (25.98)^2/10 = 67.5 m (c) at t = 2 sec rate of change in speed = 1.89 m/s^2 rho = v^2/g cos theta prime = 71.28 m