Suzy drops a rock from the roof of her house.Mary sees the rock pass her 3.3 m tall windowin 0.142 sec.From how high above the top of the windowwas the rock dropped? The acceleration ofgravity is 9.8 m/s^2.Answer in units of m
Another method that gives the same answer as the two above is to use s = 0.5 * g * t^2 Let the distance to the top of the window be s0 and the time to reach it be t0. Then: s0 = 0.5 * g * t0^2 and s0 + 3.3 = 0.5 * g * (t0 + 0.142)^2 from which, by subtraction: 6.6 / g - 0.142^2 = 0.284 * t0 and so t0 = 2.300371~ seconds and s0 = 25.929~ metres
Vavg for rock during its pass by the 3.3 m window = distance/time = 3.3/0.142 Vavg = 23.24 m/s Vavg = (Vo + Vf)/2 {Vo = initial speed at top of window, Vf = final speed at bottom} (2)(23.24) = Vo + Vf 46.48 = Vo + Vf Vf = Vo + gt = Vo + (9.81)(0.142) = Vo + 1.39 46.48 = Vo + Vo + 1.39 45.09 = 2Vo Vo = 22.5 m/s time for rock to fall to top of window = t = 22.5/g = 22.5/9.81 = 2.30 s distance above top of window from which rock was dropped = h = 1/2gt? h = (0.5)(9.81)(2.30)? = 25.9 m ANS
acceleration due to gravity is 9.8 m/s^2 it means that it takes one second to travel 9.8 m from rest during free fall. so 9.8m in 1 sec how many meters in 0.142 sec = 1 = 9.8 = 0.142 = ? = 0.142*9.8 = 1.396 m the rock was dropped from a height of 1.396 m above the top of window
Whilst I'm a tad unsure of what the question is actually seeking I can tell you the above/below answer is definitely WRONG. dy = y0 + v0.t +1/2a.t^2 So if a = g = -9.8m/s^2 and t = 1s then dy = 4.9m NOT 9.8m. The velocity will be 9.8m/s on the other hand though. I'll give it a stab. Clears 3.3m Window in .142s, so we need to solve for v0 and then use that to find the height required. v0 = -22.54363662m/s, or -22.543... downwards. Now that we have that, we can use the formula Vf^2 = Vo^2 +2as (the above value will be Vf here as it's it the velocity after dropped, and Vo will be 0 and it's dropped from rest) this gives that s = 25.9m above the window.
