Question

A slingshot launches a stone vertically with an initial velocity of 400 ft/s from an initial height of 10 ft. find the average velocity over time interval [3, 8]. (assume g = 32 ft/s2) ft/s

Answer 1

576 ft/s The distance the stone will have traveled is expressed by the formula d = VT + 0.5AT^2 where d = distance V = initial velocity T = time A = acceleration Since the stone is given an initial velocity of 400 ft/s, it will take 400/32 = 12.7 seconds for it to stop and start falling to the ground, which is comfortably larger than the upper time value for the interval [3,8]. So let's see how far the stone has traveled at T=3 and T=8 d = VT + 0.5AT^2 d = 400T + 16T^2 d = 400*3 + 16*3^2 = 1200 + 16*9 = 1200 + 144 = 1344 d = 400*8 + 16*8^2 = 3200 + 16*64 = 3200 + 1024 = 4224 So the stone traveled a distance of 4224-1344 = 2880 feet during the specified interval. And since the interval spanned 8-3 = 5 seconds, the average velocity will be 2880/5 = 576 ft/s