An inquisitive physics student and mountain climber climbs a 49.0 m cliff that o

Question

An inquisitive physics student and mountain climber climbs a 49.0 m cliff that overhangs a calm pool of water. He throws (not drops) two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 1.95 m/s downward. Hint: Because the motion is only in the downward direction, choose the positive y-axis to point downward.

(a) How long after release of the first stone do the two stones hit the water?

(b) What initial velocity must the second stone have if they are to hit simultaneously?
(downward)
(c) What is the speed of each stone at the instant the two hit the water?

Explanation / Answer

part a

as given in question that they releae after 1 sec from top

time taken to hit water will be same because intial speed is same only difference of 1 sec is came from starting

so ans t=1 sec

part b

now time taken by fisrt

by kinematics

h=ut+0.5gt2

49=1.95t +0.5*9.81t2

4.905t2 +1.95t -49=0

t=2.96813 , -3.365

so consider t=2.96813sec

now if we wnat to hit the water at same time by both then second should hit in( t-1) sec

so time taken by second t2 =t-1=1.96813sec

45=vi2 *1.698133 + 0.5*9.81*1.9681332

v2i =15.3111 m/s

part c

v1f = vi1 + gt=1.95 +9.81*2.968133 =31.0673m/s

v2f =vi2 + gt=15.3111 +9.81*1.968133 =34.6185m/s

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