A rock climber stands on top of a 57 m -high cliff overhanging a pool of water.

Question

A rock climber stands on top of a 57 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The initial speed of the first stone was 1.7 m/s . In all parts, use the standard sign convention for direction. a) How long does it take the first stone to hit the water? b)What was the initial velocity of the second stone? c) What is the velocity of the first stone as it hits the water? d)What is the velocity of the second stone as it hits the water?

Explanation / Answer

a)

Use kinematic eqn,

y=vi*t+1/2*at^2

Plug values for first stone,

-57.0=-1.7*t-1/2*9.8*t^2

t= 3.24s

b)

Time for 2nd stone = time for 1st stone + 1.0s

T=t-1.0

T=3.24-1.0

T=2.24s

To calculate initial velocity of 2nd stone use kinematic eqn,

y=vi*t+1/2*aT^2

Plug values for 2nd stone,

-57.0=vi*2.24-1/2*9.8*2.24^2

vi= -14.47m/s

c)

Use kinematic eqn,

vf=vi+at

Plug values for first stone,

vf= -1.7-9.8*3.24

vf= -33.45m/s

d)

Use kinematic eqn,

vf=vi+aT

Plug values for 2nd stone,

vf= -14.47-9.8*2.24

vf= -36.42m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.