physics

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explorer

2008-09-13 7:23 pm

A ball,as shown in figure is profected with an initial speed of 30m/s at an angle of 60^o above the horizontal . the ball hits the top of the building of height h, 5s after launching.

a. determine the height h of the building

b.what is the speed of the ball just before hitting on the building

c.find out the maximum height H reached above the ground.

更新1:

in horizontal, v=(30)(cos60)唔係不變GE咩?? 點解B又要搵

回答 (1)

天同

2008-09-13 7:49 pm

✔ 最佳答案

(a) Consider the vertical components,
u = 30.sin(60) m/s, t = 5 s, a = -g(=-10 m/s2), s = h
using s = ut + (1/2).at^2
h = 30.sin(60) x 5 + (1/2)(-10)(5x5) m = 4.9 m

(b) u = 30.sin(60) m/s, t = 5s, a = -10 m/s2, v=?
using v = u + at = 30.sin(60) - 10x5 m/s = -24 m/s

hence, speed = square-root[24^2 + (30.cos(60))^2] m/s

(c) u = 30.sin(60) m/s, a = -10 m/s2, v = 0 m/s, s = H
using v^2 = u^2 + 2.a.s
0 = [30.sin(60)]^2 + 2.(-10).H
hence, H = [30.sin(60)]^2/20 m

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