How do I do this physics problem? I tried everything but, i'm not getting the right answer.? - climber stands
A climber stands on a cliff 25.0 m, which overlooks a pool of calm water. He throws two stones vertically apart down to 1.00 s and pointed out that lead a unique touch. The first stone had an initial velocity of -1.40 m / s.
(a) How long after the publication of the first stone for the two stones into the water?
Wednesday, January 13, 2010
Climber Stands How Do I Do This Physics Problem? I Tried Everything But, I'm Not Getting The Right Answer.?
4 comments:
given
d = 25 m, v '= -1.40 m / s
Note that the initial velocity v 'is negative, because its downward direction ...
just use the formula ...
v't = d + 1 / 2 GT ^ 2
Note that, since the distance "below" the cliff ... D would be negative ... g is negative, it will also decrease.
So ...
-25 = -1.40 * T - 1 / 2 (9.8) * t ^ 2
Simplify ..
4.9 * t ^ 2 + 1.4 * t - 25 = 0
This is just a quadratic equation easily solvable by using the formula of second-degree ...
Solution t. ..
t = -1.4 + / - squareRoot (1.4 ^ 2 to 4 * 4.9 * (-25)) / 2 * (4.9)
t = -1.4 + / - squareRoot (1.96 + 490) / 9.8
t = -1.4 + / - 22.18 / 9.8
t = 2.12 sec ..
Information on the second stone is irrelevant here.
s = 25 m
u = - 1.4 m / s, s = ut + 4.9t ^ 2
25 =- 1.4t + 4.9t ^ 2
4.9t ^ 2-1.4t -25 = 0
t = 2.41 s
1/2gt x ^ 2 = secend Stone
if g = 10 m / s ^ 2
25 = 1 / 2 * 10 * (t2) ^ 2
50 = 10t2 ^ 2 = 5 (t2) ^ 2
T2 = sqrt5s = 2.23
Cornerstone for the
1/2vt1 x = v = V0 + GT1
v-10T1 = 1, 4
25 = 1 / 2 (10T1-1, 4) T1
10 (t1) ^ 2-1.4t1-50 = 0
T1 = 2.30 for
2.3 Secound after two stones in the water taken
UT = 0.5 s ^ 2
25 = 1.4 t 0.5 (9.81) t ^ 2
2.12sot =- t S = 2.4 (rejected)
Time = 2.12
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