1. The ceiling of a classroom is 3.75m above the floor. A student tosses an apple vertically upward, releasing it 0.50 meters above the floor what is the maximum magnitude of the initial velocity that can be given to the apple if it is not to touch the ceiling?
2.The acceleration due to gravity at the surface of Mars is roughly 4m/s2. If an astronaut on Mars were to toss a wrench upward with a speed of 10 m/sec. find (a) how long would it rise? (b) how high would it go? (c) its speed at t=3sec. and (d) its displacement at t=3sec
3.A small stone is dropped from a height of 2m above the ground (a)How long will it take to reach the ground?
(b)What is its velocity immediately before hitting the ground?
4.suppose that a hot air ballonist drops an apple over the side while the balloon is accelerating upward at 4.0 m/s2 during a lift off. (a)What is the apple's acceleration once it has been released? (b)if the velocity of the balloon is 2 m/s upward at the instant of release what is the apple's veloctiy just then?
I will be happy to critique your thinking on these. All are using the same equations.
isa isa lang wag ma sama....
1. To find the maximum magnitude of the initial velocity of the apple, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, because the apple should not touch the ceiling)
u = initial velocity (what we're looking for)
a = acceleration due to gravity (-9.8 m/s^2, as the apple is thrown upward)
s = displacement (distance between the release point and the ceiling, which is 3.75 m)
Rearranging the equation, we have:
u^2 = v^2 - 2as
Plugging in the values, we get:
u^2 = 0 - 2(-9.8)(3.75)
u^2 = 2(9.8)(3.75)
u^2 = 73.5
u ≈ √73.5
u ≈ 8.57 m/s
Therefore, the maximum magnitude of the initial velocity that can be given to the apple is approximately 8.57 m/s.
2. (a) To find how long the wrench would rise, we can use the equation:
v = u + at
Where:
v = final velocity (0 m/s, as the wrench reaches its maximum height and starts to fall)
u = initial velocity (10 m/s)
a = acceleration due to gravity (-4 m/s^2, as the wrench is tossed upward on Mars)
t = time (what we're looking for)
Rearranging the equation, we have:
t = (v - u) / a
t = (0 - 10) / -4
t = 2.5 seconds
Therefore, the wrench would rise for 2.5 seconds.
(b) To find how high the wrench would go, we can use the equation:
s = ut + (1/2)at^2
Where:
s = displacement (what we're looking for)
u = initial velocity (10 m/s)
a = acceleration due to gravity (-4 m/s^2)
t = time (2.5 seconds)
Plugging in the values, we get:
s = (10)(2.5) + (1/2)(-4)(2.5)^2
s = 25 - 25
s = 0 meters
Therefore, the wrench would not go any higher and reach a maximum height of 0 meters.
(c) To find the speed of the wrench at t=3 seconds, we can use the equation:
v = u + at
Where:
v = final velocity (what we're looking for)
u = initial velocity (10 m/s)
a = acceleration due to gravity (-4 m/s^2)
t = time (3 seconds)
Plugging in the values, we get:
v = 10 + (-4)(3)
v = 10 - 12
v = -2 m/s
Therefore, the speed of the wrench at t=3 seconds is 2 m/s downward.
(d) To find the displacement of the wrench at t=3 seconds, we can use the equation:
s = ut + (1/2)at^2
Where:
s = displacement (what we're looking for)
u = initial velocity (10 m/s)
a = acceleration due to gravity (-4 m/s^2)
t = time (3 seconds)
Plugging in the values, we get:
s = (10)(3) + (1/2)(-4)(3)^2
s = 30 - 18
s = 12 meters
Therefore, the displacement of the wrench at t=3 seconds is 12 meters downward.
3. (a) To find how long it will take for the stone to reach the ground, we can use the equation:
s = ut + (1/2)gt^2
Where:
s = displacement (distance between the height and the ground, which is 2 m)
u = initial velocity (0 m/s, as the stone is dropped)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (what we're looking for)
Plugging in the values, we get:
2 = 0 + (1/2)(-9.8)t^2
2 = -4.9t^2
t^2 = -2 / -4.9
t^2 ≈ 0.408
t ≈ √0.408
t ≈ 0.639 seconds
Therefore, it will take approximately 0.639 seconds for the stone to reach the ground.
(b) To find the velocity of the stone immediately before hitting the ground, we can use the equation:
v = u + gt
Where:
v = final velocity (what we're looking for)
u = initial velocity (0 m/s, as the stone is dropped)
g = acceleration due to gravity (-9.8 m/s^2)
t = time (0.639 seconds, as found in the previous question)
Plugging in the values, we get:
v = 0 + (-9.8)(0.639)
v ≈ -6.28 m/s
Therefore, the velocity of the stone immediately before hitting the ground is approximately 6.28 m/s downward.
4. (a) The apple's acceleration once it has been released is the same as the acceleration of the balloon, which is 4.0 m/s^2 upward (given in the question).
The acceleration of the apple remains the same even after it has been released.
(b) To find the apple's velocity just when it is released, we can use the equation:
v = u + at
Where:
v = final velocity (what we're looking for)
u = initial velocity (2 m/s, upward)
a = acceleration (4.0 m/s^2, upward)
t = time (0 seconds, as it is released instantaneously)
Plugging in the values, we get:
v = 2 + 4(0)
v = 2 m/s
Therefore, the apple's velocity just when it is released is 2 m/s upward.