A rock climber stands on top of a 50-m-high cliff overhanging a pool of water. He throws stones vertically downward 1.0 s apart and observes that the cause a single splash. The initial speed of the first stone was 2.0 m/s. a. How long after the release of the first stone does the second stone hit the water? b. What was the initial speed of the second stone? c. What is the speed of each stone as it hits the water?

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A rock climber stands on top of a 50-m-high cliff overhanging a pool of water. He throws stones vertically downward 1.0 s apart and observes that the cause a single splash. The initial speed of the first stone was 2.0 m/s.   a. How long after the release of the first stone does the second stone hit the water?   b. What was the initial speed of the second stone?   c. What is the speed of each stone as it hits the water?

boronganfh · Accepted Answer

Step 1  We will take downward is the positive direction of moving:  For the first stone, we can find its final velocity just before hitting the water.  The givens for the first stone:  vi=2ms  yf=50m  yi=0m because our origin is the upper point.  a=9.8ms2 Notice that the acceleration of gravity we are putting it positive because we considered the downward is positive. The velocity is increasing, So both g and v must have the same sign.  Then we will use yf=yi+vit+0.5at2 to find the time.  50=0+2t+0.5×9.8t2  4.9t2+2t−50=0  t=−b±b2−4ac2a  t=−2±22−(4×4.9×−50)2×4.9  t=(2.996,−3.4)s we reject the negative t  So tFor the first stone=3.0s  Step 2  The second stone had released after the first by one second and they reached the bottom at the same time.  So the time for the second stone is 2s.  tFor the second stone=2.0s  Step 3  The initial speed for the second one:  yf=yi+vit+0.5at2  50=0+(vi×2)+(0.5×9.8×22)  vi=50−19.62=15.2ms  vi for the second stone=15.2ms  Step 4  vf1=vi1+a t1  vf1=2+(9.8×3)=31.4ms  Step 5  vf2=vi2+a t2  vf2=15.2+(9.8×2)=34.8ms

Neunassauk8 · Accepted Answer

Step 1  The downward id the positive direction or moving  For the first stone  We can find the final velocity just before hitting the water  The given for the first stone  vi=2ms  yf=50m  yi=0m  Thus our origin is the upper point  Step 2  a) The acceleration of gravity we are putting it positive because we considered the downward is positive the velocity is increasing  So both g and v must have the same sign  yf=yi+vi+0.5at2  50=0+2t+0.5×9.8t2  4.92+2t−50=0  t=−b±b2−4ac2a  =−222−(4×4.9×−50)2×4.9  t=(2.996−3.4)s  Thus t(for the first stone) =3.0s  The second stone had released after the first by one second and they reached the bottom at the same time  So the time for the second stone is 2s  Step 3  b) The initial speed of the second stone  yf=yi+vi+0.5at2  50=0+(vi×2t+0.5×9.8×22)  vi=50−19.62  =15.2ms  c) vi for the second stone is 15.2m/s  vf1=vi1+at1  =2+(9.8×3)  =31.4ms  vf2=vi2+at2  =15.2+(9.8×2)  =34.8ms

karton · Accepted Answer

For the first stone: The distance travelled is the initial velocity times time + distance for an acceleration from 0. In this case, the acceleration is g (the acceleration of gravity 9.8m/s2) So d=vi⋅t+12⋅g⋅t2 Plug in the numbers: 50=2⋅t+9.82⋅t2 Put in standard format: 4.9t2+2t−50=0 not obviously factorable, so use the quadratic formula: t=(−2±22−4⋅4.9⋅−502(4.9)) t=−2±9849.8 t=−2±31.49.8 A negative value has no meaning so t=29.49.8=3 (This is the (a) answer) This is the time after the 1st throw that BOTH stones hit the water, so the time for the second stone is 2 seconds. Use the same equations to get the initial velocity of the 2nd stone. (This time, you&#039;ll be entering d and t and solving for vi ) For the velocity hitting the water, use v=vi+g⋅t=vi+9.8t plug in vi and t, and solve

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