Physics Problem using Equations of Motion

In summary: For (d), the vertical axis is the velocity, the horizontal axis is the time. Draw the axes, label them with the corresponding units. The brick is released with velocity -5 m/s, it accelerates downwards and reaches the ground with velocity 24.76 m/s. The velocity is zero at t= 2.02 s. So the line starts from (-5,0) and reaches (24.76, 0) at t=2.02 s. It is a downward parabola. ehildIn summary, a brick is dropped from a hot air balloon ascending at 5 ms^-1. The height of the balloon above the ground at the time of release was 30m. By using
  • #1
Scarecrow13
5
0

Homework Statement



A brick is dropped from a hot air balloon which is ascending at 5ms^-1. The height of the balloon above the ground at the time of release of the brick was 30m.

a. Determine the time that it took the brick to hit the ground.
b. At what height was the balloon above the ground at the instant the brick hit the ground?
c. What was the velocity of the brick when it hit the ground?
d. Sketch a graph of the velocity against time for the brick over this journey.

Homework Equations



1. v=u+at
2. v^2=u^2+2as
3. s=ut+1/2at^2
4. v=s/t

The Attempt at a Solution



For this question I specified up direction as negative and down direction as positive.

For (a)

I decided to use the formula, s=ut+1/2at^2, where I substituted s for 30, u for -5 and a for 9.8. When rearranged this gave me 4.9t^2-5t-30=0. Using the quadratic formula I then got the answer t=-2.53 and 2.53.

I would have thought that these values would have been different however as after being released from the balloon, the brick would surely have risen slightly before gravity accelerates it towards ground meaning the positive value should be greater than the negative value.

For (b)

I used the s=ut+1/2at^2 formula with s=30, u=-5 and a=0 values of the balloon to prove that after six seconds the balloon was 30m above the ground. I got a t=-6 answer for this though and thus maybe I did something wrong, it should be positive right? The I used the s=ut+1/2at^2 formula using -8.53 (-6-2.53)*-5 to give me the displacement 42.65m.

For (c)

I used v=u+at with the values, u=-5 a=9.8 and t=-2.53 to give me the velocity 19.794 ms^-1 down.

I will attempt (d) after finishing a,b and c.
 
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  • #2
Scarecrow13 said:

Homework Statement



A brick is dropped from a hot air balloon which is ascending at 5ms^-1. The height of the balloon above the ground at the time of release of the brick was 30m.

a. Determine the time that it took the brick to hit the ground.
b. At what height was the balloon above the ground at the instant the brick hit the ground?
c. What was the velocity of the brick when it hit the ground?
d. Sketch a graph of the velocity against time for the brick over this journey.

Homework Equations



1. v=u+at
2. v^2=u^2+2as
3. s=ut+1/2at^2
4. v=s/t

The Attempt at a Solution



For this question I specified up direction as negative and down direction as positive.

For (a)

I decided to use the formula, s=ut+1/2at^2, where I substituted s for 30, u for -5 and a for 9.8. When rearranged this gave me 4.9t^2-5t-30=0. Using the quadratic formula I then got the answer t=-2.53 and 2.53.

I would have thought that these values would have been different however as after being released from the balloon, the brick would surely have risen slightly before gravity accelerates it towards ground meaning the positive value should be greater than the negative value.

These results are wrong. How do you solve a quadratic equation?

The formula for the displacement you used is correct, and it accounts for the small intial rise of the brick. Try to find the displacement at t=0.1 s.

Scarecrow13 said:
For (b)

I used the s=ut+1/2at^2 formula with s=30, u=-5 and a=0 values of the balloon to prove that after six seconds the balloon was 30m above the ground. I got a t=-6 answer for this though and thus maybe I did something wrong, it should be positive right? The I used the s=ut+1/2at^2 formula using -8.53 (-6-2.53)*-5 to give me the displacement 42.65m.

The balloon was at 30 m height when the brick was dropped out. You can consider that instant as zero of the time. The balloon ascends with 5/s speed so it rises by how many m-s during the time the brick falls? What will be the final height?
But you need the correct time of fall first.

ehild
 
  • #3
I believe I solved for time wrong.

There should be a way to solve for t using the equations I listed, without using the quadratic formula.

Why solve for t=0.1? It equals -30.451.
 
  • #4
Scarecrow13 said:
I believe I solved for time wrong.

There should be a way to solve for t using the equations I listed, without using the quadratic formula.

Why solve for t=0.1? It equals -30.451.

You have to solve the quadratic equation to get the time. Use the quadratic formula or completing the square - you certainly have learned some of them?

To find the height at t=0.1 just illustrates that the formula s = ut+a/2t2 works both for positive and negative displacements.

ehild
 
  • #5
I have decided to change it to make up positive and down negative now. Makes more sense to me now.

For (a) I just realized this could also have been done by using v^2=u^2+2as to find v=24.74ms^-1 and then using v=u+at, to find t, being -2.02

For (b) Using s=ut+1/2at^2 with s=5(6+2.02) I get s=40.1m for the height of the balloon the instant the brick hit the ground.

For (c) The answer is the v=24.76ms^-1 I previously found.

Now how would I set out the axis for (d)??
 
Last edited:
  • #6
Scarecrow13 said:
I have decided to change it to make up positive and down negative now. Makes more sense to me now.

For (a) I just realized this could also have been done by using v^2=u^2+2as to find v=24.74ms^-1 and then using v=u+at, to find t, being -2.02

For (b) Using s=ut+1/2at^2 with s=5(6+2.02) I get s=40.1m for the height of the balloon the instant the brick hit the ground.

For (c) The answer is the v=24.76ms^-1 I previously found.

Now how would I set out the axis for (d)??

You are right you can find the time without the quadratic formula, but your result is not correct. Is it possible that the brick reaches the ground in negative time? Before it has been dropped out from the balloon?
Velocity has sign, positive when up and negative when moving down. Is that v=24.74 m/s upward or downward? Positive or negative?

ehild
 
  • #7
It is 24.76ms^-1 downwards.
 
  • #8
What values you have to plug in for u and v into the equation v=u+at?

ehild
 
  • #9
If you rearrange v=u+at, you get t=(v-u)/t

I substituted 24.76 for v, and 5 for u.

Just while typing this I realize that I should be substituting -24.76 and thus getting t=3.03.

This will thus change my answers for (a) and (b). Thanks for calling my attention to that.
 
  • #10
Scarecrow13 said:
Now how would I set out the axis for (d)??

Choose t (time) for the horizontal axis and v (velocity) as vertical.

ehild
 

1. What are the three equations of motion in physics?

The three equations of motion in physics are:

1. v = u + at (final velocity = initial velocity + acceleration x time)

2. s = ut + 1/2at^2 (displacement = initial velocity x time + 1/2 x acceleration x time^2)

3. v^2 = u^2 + 2as (final velocity^2 = initial velocity^2 + 2 x acceleration x displacement)

2. What is the difference between speed and velocity?

Speed is the rate at which an object moves, while velocity is the rate at which an object moves in a specific direction. Speed is a scalar quantity, while velocity is a vector quantity that includes both magnitude and direction.

3. How are displacement and distance different?

Displacement is the change in position of an object from its starting point to its end point. It is a vector quantity that includes both magnitude and direction. Distance, on the other hand, is the total length of the path traveled by an object. It is a scalar quantity and does not take into account the direction of motion.

4. What is the acceleration due to gravity?

The acceleration due to gravity is approximately 9.8 m/s^2 (meters per second squared) near the Earth's surface. This means that an object in free fall will accelerate at a rate of 9.8 meters per second squared towards the ground.

5. How do you use the equations of motion to solve physics problems?

To solve a physics problem using the equations of motion, you first need to identify the known values and the unknown value. Then, you can plug in the known values into the appropriate equation and solve for the unknown value. It is important to pay attention to units and use the correct equation for the given scenario.

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