Solving Projectile Problem: Initial Speed 100m/s, Object 5m Away 2m High

  • Thread starter Mazero
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In summary, the conversation is about trying to find the angle of an object launched at an unknown angle with an initial speed of 100m/s, in order to clear an object that is 5m away and 2m high. The person has made several attempts using equations such as y=(Vy + (1/2)(-9.8)(t^2) and Vx^2 = 100 - Vy^2, but has not been successful. They suggest trying to solve for theta using the equation Y = f(x) and plugging in the given values.
  • #1
Mazero
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Im given the initial speed (100m/s) of an object launched at an unkown angle, it needs to just barely go over an object 5m away 2m high.

Ive made several attempts at this, all of which ending up in the trash
my most recent try is

y=(Vy + (1/2)(-9.8)(t^2)
t = 5/(Vx)

Then used Vx^2 = 100 - Vy^2
and subsituted it all in, and i come out with a big mess to solve, so I've assumed this is not the correct (or best) way to do this

Sorry for not going into more detail, but you should be able to get the idea
 
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  • #2
BTW I am trying to find the angle...and the 100m/s is the Hypot. componet of the velocity
 
  • #3
try putting Y = f(x), so you can plug your X, Y and Vo values, and solve for theta.
 

1. How do I find the angle of projection for a projectile with an initial speed of 100m/s and an object 5m away and 2m high?

To find the angle of projection, you can use the formula: θ = tan-1 (h/5), where θ is the angle of projection, h is the height of the object, and 5 is the horizontal distance. In this case, θ = tan-1 (2/5) = 22.6°.

2. What is the maximum height reached by the projectile?

The maximum height reached by the projectile can be calculated using the formula: H = (v0^2 * sin^2(θ))/2g, where v0 is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity (9.8m/s^2). In this case, H = (100^2 * sin^2 (22.6°))/19.6 = 51.4m.

3. How long does it take for the projectile to reach the object?

The time taken for the projectile to reach the object can be calculated using the formula: t = (2 * v0 * sin(θ))/g, where v0 is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity (9.8m/s^2). In this case, t = (2 * 100 * sin(22.6°))/9.8 = 10.4s.

4. What is the total horizontal distance covered by the projectile?

The total horizontal distance covered by the projectile can be calculated using the formula: d = v0 * cos(θ) * t, where v0 is the initial speed, θ is the angle of projection, and t is the time taken. In this case, d = 100 * cos(22.6°) * 10.4 = 945.7m.

5. How can I determine the final velocity of the projectile when it reaches the object?

The final velocity of the projectile can be calculated using the formula: v = √(v0^2 + 2gh), where v0 is the initial speed, g is the acceleration due to gravity, and h is the height of the object. In this case, v = √(100^2 + 2*9.8*2) = 104.2m/s.

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