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??
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.
@kunoi i didn't got f, g &h
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.
To solve this problem, we'll use the equations of motion and the principles of conservation of energy.
(a) The speed of stone 1 at the bottom of the cliff can be found using the equation of motion for uniformly accelerated motion:
v^2 = u^2 + 2as
Here, v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
For stone 1, it is thrown vertically downwards, so the initial velocity u will be positive 10 m/s, acceleration a will be the acceleration due to gravity (-10 m/s^2), and the distance s will be the height of the cliff (24 m).
Substituting these values into the equation, we get:
v^2 = (10 m/s)^2 + 2*(-10 m/s^2)*24 m
v^2 = 100 m^2/s^2 - 480 m^2/s^2
v^2 = -380 m^2/s^2
Since we cannot have a negative value for speed, we take the positive square root:
v = √(380 m^2/s^2) ~ 19.5 m/s
So, the speed of stone 1 at the bottom of the cliff is approximately 19.5 m/s.
To answer the rest of the questions, we'll use the principle of conservation of energy.
(b) Since stone 2 is thrown vertically upwards at the same speed, it will reach the same height as the cliff before eventually falling back down. Therefore, its speed at the bottom of the cliff will be the same as stone 1, i.e., approximately 19.5 m/s.
(c) To find the time of flight of stone 1, we can use the equation of motion for vertical motion:
s = ut + (1/2)at^2
Here, s is the distance (height of the cliff), u is the initial velocity, a is the acceleration, and t is the time of flight.
Since the stone is thrown vertically downwards, u will be positive 10 m/s, a will be the acceleration due to gravity (-10 m/s^2), and s will be the height of the cliff (24 m).
Plugging in these values, we get:
24 m = 10 m/s * t + (1/2) * (-10 m/s^2) * t^2
Rearranging the equation, we have:
5t^2 - 10t - 24 = 0
Solving this quadratic equation, we find two solutions for t: t = 2.2 s (rounded to one decimal place) and t = -2.2 s. Since time cannot be negative, we discard the negative value.
Therefore, the time of flight of stone 1 when it hits the bottom of the cliff is approximately 2.2 s.
(d) Stone 2 follows the same trajectory as stone 1, but in reverse. So, the time of flight for stone 2 will also be approximately 2.2 s.
(e) The average speed of stone 1 during its flight can be calculated as the total distance traveled divided by the total time taken:
Average Speed = Total Distance / Total Time
For stone 1, the total distance traveled is equal to the height of the cliff (24 m), and the total time taken is the time of flight (2.2 s).
Average Speed = 24 m / 2.2 s ~ 10.9 m/s
Therefore, the average speed of stone 1 during its flight is approximately 10.9 m/s.
(f) The average speed of stone 2 during its flight will also be approximately 10.9 m/s since it travels the same distance in the same time as stone 1.
(g) The magnitude of average velocity is given by the total displacement divided by the total time taken:
Average Velocity = Total Displacement / Total Time
For stone 1, the total displacement is the distance from the top of the cliff to the bottom, which is equal to the height of the cliff (24 m). The total time taken is the time of flight (2.2 s).
Average Velocity = 24 m / 2.2 s ~ 10.9 m/s
Therefore, the magnitude of the average velocity of stone 1 during its flight is approximately 10.9 m/s.
(h) The magnitude of the average velocity of stone 2 during its flight will also be approximately 10.9 m/s since it travels the same displacement in the same time as stone 1.
I hope this helps! Let me know if you have any further questions.