Question

An inquisitive physics student and mountain climber climbs a 49.0-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 1.80 m/s.

(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 the two stones are to hit the water simultaneously?

magnitude	m/s
direction

(c) What is the speed of each stone at the instant the two stones hit the water?

first stone	m/s
second stone	m/s

Answer 1

here,

h = 49 m

a)

for first stone

initial velocity , u1 = 1.8 m/s

let the time taken to hit water be t1

h = u1 * t1 + 0.5 * g * t1^2

49 = 1.8 * t1 + 0.5 * 9.81 * t1^2

solving for t1

t1 = 2.98 s

b)

t2 = t1 - 1 s = 1.98 s

let the initial velocity of seccond stone be u2

h = u2 * t2 + 0.5 * g * t2^2

49 = u2 * 1.98 + 0.5 * 9.81 * 1.98^2

solving for u2

u2 = 15 m/s

direction is downwards

c)

the final speed of stone 1 , v1 = u1 + g * t1

v1 = 1.8 + 9.81 * 2.98 m/s = 31 m/s

the final speed of seccond stone , v2 = u2 + g * t2

v2 = 15 + 9.81 * 1.98 m/s

v2 = 34.4 m/s