f.4 phys..

用戶97935

2008-10-16 7:34 am

A stone is projected from the ground vertically upwards. It reaches the maximum height and falls back to the ground. At the instant moment, the stone falls to the same height as the upper edge of the window. It takes 0.16s for the stone to pass the window whose height is 1.5m. Suppose the lower edge of the window is 1.2m above the ground. Assume that air resistance is negligible.

a) Find the initial velocity at which the stone is projected. Take upward direction to be positive.

b) What is the maximum height reached by the stone?

回答 (1)

Uncle Michael

2008-10-16 12:01 pm

✔ 最佳答案

a)
The stone goes up and falls to the same height as the upper edge of the window:
u = ?
s = 1.2 + 1.5 = 2.7 m
t = t1
a = -10 m s-2

s = ut + (1/2) at2
2.7 = ut1 + (1/2)(-10)t12
ut1 - 5t12 = 2.7 ...... (1)

The stone goes up and falls to the same height as the lower edge of the window:
u = ?
s = 1.2 m
t = t1 + (0.16 s)
a = -10 m s-2

s = ut + (1/2) at2
1.2 = u(t1 + 0.16) + (1/2)(-10)(t1 + 0.16)2
u(t1 + 0.16) - 5(t1 + 0.16)2 = 1.2 ...... (2)

(2) - (1):
0.16u - 5[(t1 + 0.16)2 - t12] = -1.5
0.16u - 5[t12 + 0.32t + 0.0256 - t2] = -1.5
0.16u - 1.6t1 - 0.128 = -1.5
1.6t1 = 0.16u + 1.372
t1 = 0.1u + 0.8575 ...... (3)

Put (3) into (1)
u(0.1u + 0.8575) - 5(0.1u + 0.8575)2 = 2.7
0.1u2 + 0.8575u - 0.05u2 - 0.8575u - 3.6765 = 2.7
0.05u2 = 6.3765
The initial speed of stone, u = 11.29 m s-1 (upwards)

=====
The stone goes up to the maximum height:
u = 11.29 m s-1
v = 0 m s-1
s = ?
a = -10 m s-2

v2 = u2 + 2as
0 = (11.29)2 + 2(-10)s
Maximum height, s = 25.49 m
=

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