Calculating Initial Velocity Needed to Reach Specified Height

In summary, the conversation discusses the problem of determining the initial velocity and velocity decrement needed for a projectile to reach a predetermined height and time, taking into account gravity as a negative force. The given equations and variables are used to calculate the initial velocity and velocity decrement, but further adjustments may be necessary to ensure the projectile reaches the specified height and time.
  • #1
goombachu
4
0

Homework Statement



I have a projectile that shoots straight up into the air (parallel to the y-axis). I need to calculate the initial velocity needed to reach a specific height. I am given the following:
- the y-displacement (s1) to reach
- the time (t) at which the projectile reaches the specified height
- the acceleration due to gravity (a = -9.8 m/s/s)

Additionally, after time (t), velocity should be zero, and the projectile should be at the specified height. In other words, it takes (t) seconds to reach the "apex."

Homework Equations



Nothing given, but I suspect:
s1 = s0 + (v0 * t) + (0.5 * a * t * t)

The Attempt at a Solution



Let's say:
- s0 = 0 meters
- s1 = 1 meters
- t = 0.25 secs
- a = -9.8 m/s/s
- v0 = ?

I thought I'd solve for v0 in the equation above. That gives:

v0 = s1 / t - (0.5 * a * t)
v0 = 1 / 0.25 - (0.5 * -9.8 * 0.25)
v0 = 4 - (-1.225)
=> v0 = 5.1225

That seems valid, but now let's say I need to update the velocity 30 times per second. This means that I am starting with an initial velocity of 5.1225 m/s, and each "tick" I need to decrement the velocity by some amount. After 0.25 seconds my velocity should tick down to zero.

I am guessing I would use (v1 = v0 + a * t) per "tick" for this, correct? Where v0 is always 5.1225 and (t) increments by (1/30) in each tick?
 
Last edited:
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  • #2


goombachu said:

Homework Statement



I have a projectile that shoots straight up into the air (parallel to the y-axis). I need to calculate the initial velocity needed to reach a specific height. I am given the following:
- the y-displacement (s1) to reach
- the time (t) at which the projectile reaches the specified height
- the acceleration due to gravity (a = -9.8 m/s/s)

Additionally, after time (t), velocity should be zero, and the projectile should be at the specified height. In other words, it takes (t) seconds to reach the "apex."

Homework Equations



Nothing given, but I suspect:
s1 = s0 + (v0 * t) + (0.5 * a * t * t)

The Attempt at a Solution



Let's say:
- s0 = 0 meters
- s1 = 1 meters
- t = 0.25 secs
- a = -9.8 m/s/s
- v0 = ?

I thought I'd solve for v0 in the equation above. That gives:

v0 = s1 / t - (0.5 * a * t)
v0 = 1 / 0.25 - (0.5 * -9.8 * 0.25)
v0 = 4 - (-1.225)
=> v0 = 5.1225 <-----------Error: =5.225


That seems valid, but now let's say I need to update the velocity 30 times per second. This means that I am starting with an initial velocity of 5.1225 m/s, and each "tick" I need to decrement the velocity by some amount. After 0.25 seconds my velocity should tick down to zero.

I am guessing I would use (v1 = v0 + a * t) per "tick" for this, correct? Where v0 is always 5.1225 and (t) increments by (1/30) in each tick?

You can use the V = Vo -a*t (treat a as negative) relationship at each 1/30 to give you V at each point leading up to 0. You can merely note in that regard that you will be decrementing by 9.8/30 = .32667m/s for each 1/30
 
  • #3


>> You can merely note in that regard that you will be decrementing by 9.8/30 = .32667m/s for each 1/30

That is what I thought as well, but:

0.25 secs * 30 ticks per second = ~8 ticks
0.32667 m/s * 8 ticks = 2.61336 meters

So my velocity doesn't tick down to zero in the 0.25 seconds I have to get there.
 
  • #4


goombachu said:
>> You can merely note in that regard that you will be decrementing by 9.8/30 = .32667m/s for each 1/30

That is what I thought as well, but:

0.25 secs * 30 ticks per second = ~8 ticks
0.32667 m/s * 8 ticks = 2.61336 meters

So my velocity doesn't tick down to zero in the 0.25 seconds I have to get there.

That's because you didn't calculate the maximum height. You derived an equation for what the initial velocity would be if after 1/4 sec it got to height 1 meter.

To calculate initial Velocity if the max height is 1 m then you would choose V2 = 2*(9.8)*1
That means V = 4.427m/s and the time to do that is x = 1/2*9.8*t2 and the time then is .4518 s
 
Last edited:
  • #5


OK maybe I should rephrase the problem then to be clear. I need to velocity to hit zero at a predetermined height (1 meter in this example), and it needs to take exactly a predetermined amount of time (.25 seconds in this example) to get to that height. A negative force (which I thought could be gravity) must act on the projectile in order to launch it at a high speed and slow it down as it gets closer to the height.

Need to figure out:

a) What velocity to launch the projectile at in order to reach a given height at a given time
b) How to decrease the velocity in each 1/30 tick in order to reach velocity = 0 at a given time

The method you suggest looks like it doesn't get me to the predetermined height in the predetermined amount of time. At first glance it would seem that I might need to change the acceleration in order to do that?
 
  • #6


goombachu said:
OK maybe I should rephrase the problem then to be clear. I need to velocity to hit zero at a predetermined height (1 meter in this example), and it needs to take exactly a predetermined amount of time (.25 seconds in this example) to get to that height. A negative force (which I thought could be gravity) must act on the projectile in order to launch it at a high speed and slow it down as it gets closer to the height.

Need to figure out:

a) What velocity to launch the projectile at in order to reach a given height at a given time
b) How to decrease the velocity in each 1/30 tick in order to reach velocity = 0 at a given time

The method you suggest looks like it doesn't get me to the predetermined height in the predetermined amount of time. At first glance it would seem that I might need to change the acceleration in order to do that?

Looks like that's all that's left that you can change then.

On a planet far, far, away there is a boy playing with a ball ...
 
  • #7


Right. So now i have 2 unknowns then: initial velocity and acceleration.

I don't readily see any of the standard kinematic equations that could help me find both? The only one that doesn't involve both unknown terms is d = avg_vel * t, but that doesn't really give me an initial "launch" velocity does it?
 

1. How do you calculate the initial velocity needed to reach a specified height?

To calculate the initial velocity needed to reach a specified height, you can use the formula v = √(2gh), where v is the initial velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the specified height.

2. What units should be used when calculating initial velocity for a specified height?

The units for initial velocity will depend on the units used for acceleration due to gravity and height. In most cases, it is recommended to use meters per second (m/s) for velocity, meters (m) for height, and meters per second squared (m/s²) for acceleration due to gravity.

3. Can the initial velocity needed to reach a specified height ever be negative?

No, the initial velocity needed to reach a specified height cannot be negative. This is because the initial velocity is the starting velocity, and it must have a positive value in order for an object to reach a specified height.

4. How does air resistance affect the calculation of initial velocity for a specified height?

Air resistance can affect the calculation of initial velocity for a specified height by slowing down the object's acceleration. This means that the initial velocity needed to reach a specified height may be higher in order to overcome the resistance and reach the desired height.

5. Is the calculation of initial velocity for a specified height affected by the mass of the object?

Yes, the mass of the object does affect the calculation of initial velocity for a specified height. Heavier objects will require a greater initial velocity to reach the same specified height compared to lighter objects, as they have more inertia and require more force to move upward.

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