Free Fall and a Projectile motion question

In summary: Here's a thing. I reckon that if ##h## is the height of the walls, ##a## is the distance to the first wall and ##b## the distance to the second wall, then:##b = \frac{ah}{a\tan \theta - h}##In this case ##a = h = 2m## so:##b = \frac{2}{\tan \theta - 1}m##Is that interesting?Yes, that is interesting! It simplifies the calculation a bit. Good job!In summary, the first problem involves finding the time and speed of a diver jumping from a height of 10m with a velocity of 8m/s. The second problem involves
  • #1
rockcandy
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Homework Statement


1. A diver jumps from a height of 10m leaving with a velocity of 8m/s find the time for which the diver is in the air and the speed at which the diver enters the water?

2. A ball is launched from point A on a horizontal plane with an angle of 55*. It passes two walls both 2m high. Wall one is 2m from point A. Find the speed of the ball and the Wall two distance from point A

The attempt at a solution
1. Vy= 8m/s dy = 10m ay=-9.8m/s2

Vfy2 = Viy2 + 2ayd

Vy=√(8m/s)2 + 2(-9.8m/s2)(-10m)
Vy = 16.1 m/s

-10 m = 8 m/s t + 1/2 (-9.8m/s2) t2
0 = -4.9 m/s2 t2 + 8m/s t + 10m

t = 2.45 s or - 0.82 s

2. Vx = VicosΘ and Vy = VisinΘ h =2m and horizontal distance = 2m

dy= Viyt + 1/2 at2 dx = Vixt + 1/2at2
since horizontal acceleration = 0 ; dx = Vixt
dy= Viyt + 1/2 at2 dx = Vixt ; 2 m = VicosΘ t
2 = VisinΘ t - 1/2 (9.8m/s2) t 2
2 = VisinΘ (2m /VicosΘ) - 1/2 (9.8m/s2)(2m /VicosΘ)2
2 = 2m tan Θ - 4.9m/s2 (12.15 m 2 /Vi2 )
2 = 2.85 m - 59.535 / Vi2
Vi = √70.04
Vi = 8.36 m/s

Im stuck on how to get the distance assuming this is correct. If I use the max range formula

x = Vi2 sin2Θ / g
x = (8.36m/s)2 sin 2(55) / 9.8 m/s2
x = 6.69 m

If I take half this distance ( 6.69m)(.5) = 3.34 m
I know that it is reaching a height of 2 m twice, and given that the parabola is symmetrical I know that on the opposite side it is also 2m away from Wall 2.
3.34 m - 2 m = 1.34 m , which the remaining distance between the wall to the half the max range.

therefore 1.34m x 2 = 2.68m which is the total distance between Wall 1 and Wall 2

So to find the distance from Wall 2 to Point A just add the total = 2.68m + 2 m = 4.68m

 
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  • #2
rockcandy said:

Homework Statement


1. A diver jumps from a height of 10m leaving with a velocity of 8m/s find the time for which the diver is in the air and the speed at which the diver enters the water?

The attempt at a solution
1. Vy= 8m/s dy = 10m ay=-9.8m/s2

Vfy2 = Viy2 + 2ayd

Vy=√(8m/s)2 + 2(-9.8m/s2)(-10m)
Vy = 16.1 m/s

-10 m = 8 m/s t + 1/2 (-9.8m/s2) t2
0 = -4.9 m/s2 t2 + 8m/s t + 10m

t = 21.6 s or -20s


Does ##21.6s## not seem a long time to be in the air?
 
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  • #3
PeroK said:


Does ##21.6s## not seem a long time to be in the air?
Youre right... my math was off
its t = 2.45s or -0.82

assuming everything else is correct
 
  • #4
rockcandy said:
Youre right... my math was off
its t = 2.45s or -0.82

assuming everything else is correct

You have to make your mind up about whether it's ##2.45s## or ##-0.82s##. Do you understand why there are two solutions?

In question 2, are you supposed to assume that the ball just clears each wall?
 
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  • #5
PeroK said:
You have to make your mind up about whether it's ##2.45s## or ##-0.82s##. Do you understand why there are two solutions?

In question 2, are you supposed to assume that the ball just clears each wall?

The time is t = 2.45 s
For #2 yes you assume it clears each wall
 
  • #6
rockcandy said:
The time is t = 2.45 s
For #2 yes you assume it clears each wall

I really like your approach to question 2. Using the range (or you could have used the distance to the highest point) and then the symmetry of the parabola was very neat.

One suggestion is to try to use algebra a little more and not be so quick to plug in the numbers.
 
  • #7
PeroK said:
I really like your approach to question 2. Using the range (or you could have used the distance to the highest point) and then the symmetry of the parabola was very neat.

One suggestion is to try to use algebra a little more and not be so quick to plug in the numbers.

Thanks ; But did I solve the 2 problems correctly?
 
  • #8
rockcandy said:
Thanks ; But did I solve the 2 problems correctly?

Yes, of course!
 
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  • #9
PS I think the simpler approach was to find the two times that ##y = 2m## and the use the x-velocity to get the distance to the second wall.
 
  • #10
PeroK said:
PS I think the simpler approach was to find the two times that ##y = 2m## and the use the x-velocity to get the distance to the second wall.
Thanks for the help , I drew a diagram and I see what you mean.
 
  • #11
rockcandy said:
Thanks for the help , I drew a diagram and I see what you mean.

Here's a thing. I reckon that if ##h## is the height of the walls, ##a## is the distance to the first wall and ##b## the distance to the second wall, then:

##b = \frac{ah}{a\tan \theta - h}##

In this case ##a = h = 2m## so:

##b = \frac{2}{\tan \theta - 1}m##

Is that interesting?
 
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FAQ: Free Fall and a Projectile motion question

1. What is the difference between free fall and projectile motion?

Free fall refers to the motion of an object when it is only under the influence of gravity, while projectile motion involves both the effects of gravity and the initial velocity of the object. In free fall, the object's velocity is constantly changing due to gravity, while in projectile motion, the object's velocity remains constant in the horizontal direction and changes only in the vertical direction due to gravity.

2. How is acceleration related to free fall?

In free fall, the acceleration is always constant and equal to the acceleration due to gravity, which is approximately 9.8 m/s². This means that the object's velocity changes by 9.8 m/s every second it is in free fall.

3. Can an object experience free fall and projectile motion at the same time?

Yes, an object can experience both free fall and projectile motion simultaneously. This occurs when an object is launched at an angle, such as in a curved trajectory, and is still under the influence of gravity.

4. How does air resistance affect free fall and projectile motion?

Air resistance, also known as drag, can have a significant effect on both free fall and projectile motion. In free fall, air resistance can slow down an object, causing it to have a lower acceleration than the acceleration due to gravity. In projectile motion, air resistance can affect the object's trajectory, causing it to deviate from its expected path.

5. What is the formula for calculating the distance traveled in free fall or projectile motion?

The formula for calculating the distance traveled in free fall or projectile motion is d = vi * t + 1/2 * a * t², where d is the distance, vi is the initial velocity, t is the time, and a is the acceleration (in the case of free fall, a is equal to the acceleration due to gravity). This formula can be used for both horizontal and vertical distances in projectile motion.

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