What is the smallest initial velocity that is required to throw a stone 49 feet up to the top of a silo?
Write down the height as a function of v_0. Then find the maximal height. (simply take the derivative and set it equal to 0 to find the time of the maximal height.) Now you have the maximal height as a function of v_0. Set this equal to 49 and solve for v_0.
u = initial velocity v = final velocity a = acceleration d = distance travelled t = time taken assumption: this is a 2-D problem. stone thrown vertically up. v2 = u2 +ad ======================== working: d = [(v+u)/2]t t = 2d/(v+u) sub into v = u +at v = u + 2ad/(v+u) multiply both sides by (v+u) v2 +uv = uv + u2 + 2ad v2 = u2 + 2ad (proved) ========================== v = 0, a = -g = -32, d = 49 0 = u2 + 2(-32)(49) u2 = 3136 u = 56 ft/s (ans)
