on the moon, the acceleration due to gravity is 1.6m/sec^2
a. if a rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30sec later?
b. how far below the point of release is the bottom of the crevasse?
c. if instead of being released from rest, the rock is thrown into a crevasse from the same point with a downward velocity of 4m/sec, when will it hit the bottom and how fast will it be going when it does?
a. Just before hitting the bottom after 30 seconds, the rock will be going, "Weeeeeeee!" Really fast! Well, technically, its velocity can be calculated using the formula: final velocity = initial velocity + (acceleration x time). So we can plug in the values and find out how fast it will be going. Ready for the calculation? Here it goes:
final velocity = 0 + (1.6m/sec^2 x 30sec) = 48m/sec
b. To calculate how far below the point of release the bottom of the crevasse is, we can use the formula: distance = 1/2 x acceleration x time^2. Plugging in the values:
distance = 1/2 x 1.6m/sec^2 x (30sec)^2 = 720m
So, the bottom of the crevasse is approximately 720 meters below the point of release.
c. If the rock is thrown into the crevasse with a downward velocity of 4m/sec, we can find out when it will hit the bottom and how fast it will be going using a similar approach as in part a.
Let's find out when it will hit the bottom first. We can use the formula: time = (final velocity - initial velocity) / acceleration. Plugging in the values:
time = (0 - (-4m/sec)) / 1.6m/sec^2 = 2.5sec
So, the rock will hit the bottom after approximately 2.5 seconds from the moment it is thrown.
Now, let's calculate the final velocity using the formula from part a:
final velocity = 4m/sec + (1.6m/sec^2 x 2.5sec) = 8m/sec
So, when the rock hits the bottom, it will be going at a speed of 8 meters per second.
To solve these problems, we can use the equations of motion with constant acceleration. Here are the step-by-step solutions for each question:
a. To calculate the velocity of the rock just before it hits the bottom, we can use the formula:
v = u + at
Where:
v = final velocity (to be determined)
u = initial velocity (0 m/s since the rock is dropped from rest)
a = acceleration due to gravity on the moon (-1.6 m/s^2)
t = time taken to reach the bottom (30 seconds)
Plugging in the values, we have:
v = 0 + (-1.6) * 30
v = -48 m/s
The negative sign indicates that the rock is moving in the opposite direction of the positive direction we initially defined. Therefore, the rock will be going 48 m/s downwards just before it hits the bottom.
b. To calculate the distance below the point of release to the bottom of the crevasse, we can use the formula:
s = ut + 0.5 * a * t^2
Where:
s = distance (to be determined)
u = initial velocity (0 m/s)
a = acceleration due to gravity on the moon (-1.6 m/s^2)
t = time taken to reach the bottom (30 seconds)
Plugging in the values, we have:
s = 0 + 0.5 * (-1.6) * 30^2
s = -720 m
The negative sign indicates that the distance below the point of release is in the opposite direction of the positive direction we initially defined. Therefore, the bottom of the crevasse is 720 meters below the point of release.
c. To calculate when the rock will hit the bottom and its velocity at that time, we can use the formula:
s = ut + 0.5 * a * t^2
Where:
s = distance (unknown)
u = initial velocity (4 m/s downwards)
a = acceleration due to gravity on the moon (-1.6 m/s^2)
t = time taken to reach the bottom (unknown)
Substituting the values given, we have:
s = 4t + 0.5 * (-1.6) * t^2
s = 4t - 0.8t^2
We want to find the time when the rock hits the bottom, so we set s = 0:
0 = 4t - 0.8t^2
Rearranging the equation, we have:
0.8t^2 - 4t = 0
Factoring out t, we have:
t(0.8t - 4) = 0
This results in two solutions: t = 0 (which is not relevant in this case since we are looking for the time it takes to hit the bottom), and t = 5 seconds.
Therefore, the rock hits the bottom of the crevasse after 5 seconds. We can calculate its velocity at this time using the equation:
v = u + at
Where:
v = final velocity (to be determined)
u = initial velocity (4 m/s downwards)
a = acceleration due to gravity on the moon (-1.6 m/s^2)
t = time taken to reach the bottom (5 seconds)
Plugging in the values:
v = 4 + (-1.6) * 5
v = 4 - 8
v = -4 m/s
Again, the negative sign indicates that the rock is moving downwards. Therefore, when it hits the bottom, it will be going 4 m/s downwards.
In order to answer these questions, we need to use the equations of motion and the principles of free fall. Let's go step by step and calculate the solutions for each question.
a. To find the speed of the rock just before it hits the bottom, we can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given:
u = 0 (since the rock is dropped)
a = -1.6 m/sec^2 (acceleration due to gravity on the moon)
t = 30 sec
Plugging in these values in the equation, we can find the final velocity:
v = 0 + (-1.6 * 30)
v = -48 m/sec
So, the rock will be traveling downward with a speed of 48 m/sec just before it hits the bottom.
b. To calculate the distance below the point of release to the bottom of the crevasse, we can use the equation:
s = ut + (1/2)at^2
where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time taken.
Given:
u = 0 (since the rock is dropped)
a = -1.6 m/sec^2 (acceleration due to gravity on the moon)
t = 30 sec
Plugging in these values in the equation, we can find the distance:
s = 0 * 30 + (1/2) * (-1.6) * (30^2)
s = -720 m
So, the bottom of the crevasse is 720 m below the point of release.
c. To find when the rock will hit the bottom and the speed at that time, we can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given:
u = -4 m/sec (velocity of the rock)
a = -1.6 m/sec^2 (acceleration due to gravity on the moon)
Plugging in these values in the equation, we can find the time taken:
v = -4 + (-1.6)t
-1.6t = -4
t = 2.5 sec
So, the rock will hit the bottom of the crevasse after 2.5 seconds. To find the speed at that time, substitute t = 2.5 sec into the equation:
v = -4 + (-1.6 * 2.5)
v = -4 + (-4)
v = -8 m/sec
Therefore, the rock will hit the bottom with a speed of 8 m/sec in the downward direction after 2.5 seconds.
vfinal=a*time
h=1/2 a time^2
h=vinitial*time+1/2 a time^2
solve the last for time, use the quadratic equation.