Calculating Time and Speed From Verticle Height and Acceleration Due to Gravity

In summary, the question is asking how to calculate the original speed and time of flight for a projectile that reaches a maximum height of 15m. To solve this, we can use the equations of motion for constant acceleration. The first equation, 2a(x-x_0)=v^2-v_0^2, can be used to find the launch speed, and the second equation, v=at+v_0, can be used to find the flight time. By understanding the concepts behind these equations, we can easily solve this problem.
  • #1
bobbles22
17
0
I have a very simple question. If I know that a projectile reaches a maximum height of 15m from its launch point, how to I work out the original speed and the time of flight given the only force is the downward acceleeration due to gravity. I'm sure its very easy, but I can';t remmeber the equation.
Many thanks
Bob
 
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  • #2
Hi there.

15m is a relatively small difference in height, so we can simplify the situation by using an approximation of a constant gravitational acceleration [tex]g[/tex].

First off, let's find the lauch speed. Are you familiar with the equations of motion for constant acceleration? One of these is [tex]2a(x-x_0)=v^2-v_0^2[/tex]. We can use this to find the launch speed. We know that the height difference is 15m meters, and that at 15m, the projectile has zero velocity. Choosing positive y-direction upwards and gravitational acceleration [tex]-9.80\frac{m}{s^2}[/tex],our equation above then becomes

[tex]2\cdot (-9.80 \frac{m}{s^2})\cdot (15m)=-v_0^2[/tex]

I'll leave you to compute the value of [tex]v_0[/tex].

Having found [tex]v_0[/tex], we can use another of the motional equation and a bit of knowledge about motion with constant acceleration to find the flight time. We know that when the projectile reaches it's atarting position again, its velocity will have equal magnitude to it's launch velocity, but opposite sign. We will us the motional equation [tex]v=at+v_0[/tex] to find the flight time. As I explained above, we want to find [tex]t[/tex] when [tex]v=-v_0[/tex]. We plug it into the equation and solve for [tex]t[/tex] and we get

[tex]t=\frac{-2v_0}{-9.80\frac{m}{s^2}}[/tex]

Again, I will leave you to compute the value of [tex]t[/tex].

The important thing is not just to know which equation to apply to a problem, but to fully understand why the equation is applicable and what it means.
 
  • #3


Hello Bob,

To calculate the original speed and time of flight of a projectile based on its maximum vertical height and acceleration due to gravity, you can use the following equations:

1. To calculate the time of flight (t):
t = √(2h/g)
Where h is the maximum height and g is the acceleration due to gravity (9.8 m/s^2 on Earth).

2. To calculate the original speed (v):
v = gt
Where g is the acceleration due to gravity and t is the time of flight calculated in step 1.

So for your specific example, if the maximum height is 15m, the time of flight would be:
t = √(2*15/9.8) = 1.76 seconds

And the original speed would be:
v = 9.8 * 1.76 = 17.25 m/s

I hope this helps! Remember, these equations only work for objects in free fall (neglecting air resistance) and on Earth's surface. Let me know if you have any further questions. Happy calculating!

Best,
 

1. How do you calculate the time and speed of an object falling from a certain height?

The time and speed of an object falling from a certain height can be calculated using the formula t = √(2h/g) and v = gt, where t is time, h is height, g is the acceleration due to gravity, and v is velocity. Simply plug in the values and solve for the desired variable.

2. What is the acceleration due to gravity?

The acceleration due to gravity, denoted as g, is a constant value of 9.8 m/s² on Earth. This means that for every second an object falls, it will gain a velocity of 9.8 meters per second.

3. Can the formula for calculating time and speed be used for any object?

Yes, the formula t = √(2h/g) and v = gt can be used for any object falling from a certain height, regardless of its mass or shape. This is because acceleration due to gravity is constant for all objects on Earth, regardless of their differences.

4. How do you convert the units for time and speed in this calculation?

To convert the units for time and speed, simply use the appropriate conversion factors. For example, if the height is given in feet, it can be converted to meters by multiplying by 0.3048. Similarly, if the velocity is given in kilometers per hour, it can be converted to meters per second by multiplying by 0.2778.

5. Is there a way to calculate the time and speed of an object if it is thrown upwards instead of falling from a certain height?

Yes, the same formula t = √(2h/g) and v = gt can be used for an object thrown upwards. In this case, the height h would be the maximum height reached by the object, and the acceleration due to gravity g would have a negative value of -9.8 m/s². The time calculated would be the time taken for the object to reach its maximum height, and the speed calculated would be the speed at which the object is thrown upwards.

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