A person is standing on the edge of a cliff of height h= 24 m. She throws a stone of mass m=0.2 kg vertically down with speed v0= 10 m/s (stone 1) and another stone of the same mass vertically up at the same speed (stone 2). The gravitational acceleration is g=10 m/s2.

Question

A person is standing on the edge of a cliff of height h= 24  m. She throws a stone of mass m=0.2 kg vertically down with speed v0= 10  m/s (stone 1) and another stone of the same mass vertically up at the same speed (stone 2). The gravitational acceleration is g=10 m/s2.

(a) What is the speed of stone 1 at the bottom of the cliff? (in m/s)

(b) What is the speed of stone 2 at the bottom of the cliff? (in m/s)

(c) What is the time of flight of stone 1 when it hits the bottom of the cliff? (in s)

(d) What is the time of flight of stone 2 when it hits the bottom of the cliff? (in s)

(e) What is the average speed of stone 1 during its flight? (in m/s)

(f) What is the average speed of stone 2 during its flight? (in m/s)

(g) What is the magnitude of the average velocity of stone 1 during its flight?
(magnitude in m/s)

(h) What is the magnitude of the average velocity of stone 2 during its flight?
(magnitude in m/s)

Please help!!How to do it??

KUNOI · Answer

Y=Y-0 +v-0*t-5t^2
solve for t (you will get a quadratic equation so use alphawolfram)
I have only solved for stone one as for stone two they havent given much detail.
then for final velocity
average speed and average velocity seem more or less the same since there is no 0 displacement and its linear motion.

ss01 · Answer

@kunoi i didn't got f, g &h

KUNOI · Answer

I didn't get h but for f and g use the formula speed=distance/time and velocity=displacement/time. Time is the same as the one you got in c and d. for distance, the first stone is 24 but for the second stone, its 24+ the upward distance calculated by the formula s=ut-5t^2. Btw the velocity of both the stones at the bottom of the cliff are the same.