A small rock is thrown vertically upward with a speed of 29.0 m/s from the edge of the roof of a 27.0-m-tall building.?
2016-02-20 6:03 am
The rock doesn't hit the building on its way back down and lands in the street below. Ignore air resistance. I got part A which was 37.0 part B, How much time elapses from when the rock is thrown until it hits the street? I don't know where to even start to get the answer.
回答 (3)
2016-02-20 7:21 am
✔ 最佳答案
Part B :Take all downward vectors to be positive.
initial velocity, u = -29.0 m/s
displacement, s = 27.0 m
acceleration, a = 9.8 m/s²
s = ut + (1/2)at²
27.0 = (-29.0)t + (1/2)(9.8)t²
4.9t² - 29.0t - 27.0 = 0
t = 6.7 or t = -0.8 (rejected)
Time taken = 6.7 s
2016-02-20 10:48 am
Divide the problem into to parts. The time taken to reach max height is u/g = 29/10 = 2.9s. And max height is u^2/2g = 29*29/2*10 =42.05m in addition to 27m of building. Total 69.05, appr 69m. Now calculate time to reach ground from this height. t=square root of 2h/g = sq root of 13.8 = 3.7s. Add both, 2.9 + 3.7 =6.6 s. The concept is correct. Check the calculations if you wish.
2016-02-20 7:18 am
h = h0 + v0*t - (1/2)gt^2 =>
0 = 27.0 + 29.0 t - 4.9 t^2;
now use the quadratic formula.
Only the positive answer will be meaningful.
It's about 7 seconds but you'll need a calculator.
0 = 27.0 + 29.0 t - 4.9 t^2;
now use the quadratic formula.
Only the positive answer will be meaningful.
It's about 7 seconds but you'll need a calculator.
收錄日期: 2021-04-18 14:35:36
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