An inquisitive physics student and mountain climber climbs a 51.0-m-high cliff t
ID: 1406073 • Letter: A
Question
An inquisitive physics student and mountain climber climbs a 51.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 1.90 m/s.
(a) How long after release of the first stone do the two stones hit the water?
_____ s
(b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?
magnitude ____s
(c) What is the speed of each stone at the instant the two stones hit the water?
Explanation / Answer
use the kinematic equation
S = So + Vo*t + 0.5 at^2 (let's take down as positive)
where So is intial time
t is time
a is accleration
so
51 = 0 + 1.90 *t + 0.5 * 9.8* t
4.9 t^2 + 1.90 t - 51 = 0
solving this quadratc equation as ax^2 + bx + c = 0
we get x = - b ± sqrt(b^2-4aC)/(2a)
so here
t = (-1.9) ± ((1.9)^2 + (4* 4.9 * 51)/(2 * 4.9)
t = -1.9 ± (31.67)/(9.8)
t = (-1.9 + 31.67) /(9.8) or (-1.9 -31.67)/9.8
t = +3.03 secs or -3.42 secs
so at t = -3.29 s, +3.03 secs
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b. then for the second stone to get there in 1 s less
51 = Vo*2.03 + 4.9(2.03)^2
Vo = (51 - 228.78)/2.03
Vo = -87.57 m/s
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2) first: v = Vo + at
V = 1.9 + (9.8 * 3.03)
V = 31.59 m/s
for second:
v = 31.59 +( 9.8 * 2.03)
V = 51.48 m/s