Magnitude of the Initial Velocity

In summary: Plugging in the known values, v= 278.0442 m/s, h= 625 m, and t= 36.4742 s, we get v= 278.0442 * 625 * 36.4742 = 1629.696 m/s, which is the maximum height the projectile will reach.
  • #1
Aerospace
39
0
Given: g = 9.8 m/s^2

A cannon sends a projectile towards a target a distance 1250m away. The initial velocity makes an angle 40 degrees with the horizontal. The target is hit.

What is the magnitude of the initial velocity?
How high is the highest point of the trajectory?
How long does it take for the projectile to reach the target?

----
I tried to find the time by using the following formula:

t = sqrt (2d/g)

and then I used

x = x(initial) +v(initial)t + 1/2gt^2

and I get a decimal answer...
And then I took the distance and divided it by 2 ad I got 625 and did tan40 = h/625 but when i entered in the answer, it's wrong.
For the 3rd Q, I used the time that I found earlier from the formula, but that's incorrect too.

Any help would be appreciated.
 
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  • #2
note that you can use the following formula, I will use 'u' for the symbol for initial velocity

v = at + u

this can model the particle's verticle component

a=-9.8
u = q sin(40) ... where q is the initial speed

if we want to find the time it takes to go up to its max point i.e. its turning point

v = 0

so, 0 = -9.8t + q sin(40)
so, t = q sin(40) / 9.8

that is the time to go up which is equal to the time to come back down again, so the total time is (I'll call the total time T)

T = 2t
T = q sin(40) / 4.9

now we revert to the horizontal component of the situation using the formula

s = s0 + u*t + 0.5*a*t^2

s0 = 0
u = q cos(40)
t = T = q sin(40) / 4.9
a = -9.8
s = 1250

so,
1250 = q cos(40) q sin(40) / 4.9 - 0.5*9.8*(q sin(40) / 4.9)^2
1250 = q^2 cos(40) sin(40) / 4.9 - q^2 sin^2(40) / 4.9
q^2 = 1250 / [ cos(40) sin(40) / 4.9 - sin^2(40) / 4.9 ]
q = √( 1250 / [ cos(40) sin(40) / 4.9 - sin^2(40) / 4.9 ] )
q ≈ 278.0442 m/s

so the initial speed of the object is approxomately 278.0442 m/s

remember that

T = q sin(40) / 4.9
T = 278.0442 sin(40) / 4.9
T ≈ 36.4742 s

so it takes approxomately 36.4742 seconds to reach the target. Now for the last part of the question we use the formula

s = s0 + u*t + 0.5*a*t^2

we will apply this to the verticle component of the situation, the time to go up to the maximum is just half of the total time of flight so

s0 = 0
t = 36.4742 / 2
u = 278.0442 sin(40)
a = -9.8

s = 278.0442 sin(40) 36.4742 / 2 - 4.9 (36.4742 / 2)^2
s ≈ 1629.696 m

so the maximum height the projectile will reach is approxomately 1629.696 m.

Thats the problem done! :smile:
 
  • #3
Your mistake was initially using "t = sqrt (2d/g)". That's the same as d= (g/2)t^2 which is the distance an object will FALL under gravity, starting from rest.

In this case, if the initial speed is v, then the initial horizontal component of speed is v cos(40) and the initial vertical component is v sin(40).

There is no acceleration horizontally and the acceleration due to gravity, -g, vertically. If we use x for horizontal distance and y for vertical distance,

x= v cos(40) t and y= v sin(40)t- g/2 t^2.

The 1250 m distance to the target is horizontal, of course, so you can solve x= v cos(40) t= 1250 to find the time, t, when the projectile hits the target. That will, of course, depend on v.

Now, use the fact that y must be 0 again to hit the target at that time: y= v sin(40)t- g/2 t^2= 0 with the value of t you just found to get an equation to solve for v.
 

What is magnitude of initial velocity?

The magnitude of initial velocity is the measure of the speed and direction of an object at the start of its motion. It is a vector quantity, meaning it has both a numerical value (magnitude) and a direction.

How is magnitude of initial velocity calculated?

Magnitude of initial velocity is calculated by determining the length of the initial velocity vector. This can be done using the Pythagorean theorem, where the magnitude is equal to the square root of the sum of the squares of the components of the vector. Alternatively, it can be calculated by multiplying the initial speed (magnitude of velocity) by the cosine of the angle between the initial velocity and the horizontal axis.

Why is magnitude of initial velocity important in physics?

Magnitude of initial velocity is important in physics because it is a crucial factor in determining the motion of an object. It affects the distance an object will travel, the time it will take to reach a certain point, and the direction it will take. It is also used to calculate other important quantities such as acceleration and momentum.

How does changing the magnitude of initial velocity affect an object's motion?

Changing the magnitude of initial velocity will result in a change in the object's speed. A larger initial velocity will result in a faster acceleration and a longer distance traveled, while a smaller initial velocity will result in a slower acceleration and a shorter distance traveled. The direction of the initial velocity will also affect the object's motion.

What factors can affect the magnitude of initial velocity?

The magnitude of initial velocity can be affected by various factors such as the force applied to the object, the angle at which the force is applied, and any external forces acting on the object (such as friction or air resistance). It can also be affected by the object's mass and its position relative to other objects.

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