An inquisitive physics student and mountain climber climbs a 51.0-m-high cliff t

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

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