Projectile fired at 30º angle

[LittleWing]

Homework Statement

Determine how far away the projectile is from the launch point when it is 50m below the launch point. Determine the max height achieved by the projectile. What is the angle that the velocity vector makes with the horizontal at the point where the projectile is -50m from launch point. V=500 m/s @ 30º angle. What is distance in horizontal and angle of velocity vector when at -50m from launch.

Homework Equations

v = u + at - Vf^2 - Vo^2 = 2aD

The Attempt at a Solution

Having trouble starting this problem...

 

[Fewmet]

I am having a little trouble following what you wrote. If I am interpreting correctly. This is a projectile launched from an elevation with an initial velocity of 500 m/s, directed 30º above the horizontal. You are asked to calculate information about the projectile when it has dropped 50 m below the launch point (and gone some distance forward at the same time).

You can approach most two-dimensional motion problems by resolving vectors into perpendicular components. In this case that means taking the 500 m/s directed up at 30º and breaking it into a vertical component and a horizontal component. You can then solve for how long it takes the projectile to rise to its peak and fall to -50 m. That is the same time the projectile moves horizontally at a constant speed (since there are no forces acting on it in the horizontal direction).

 

[miniradman]

well, you need to take into account the "speed" of gravity and how much it will slow the velocity down every second.

Putting the into in a graph will help, or even better... put the info in a graphics calculator and it will calculate the max hight ONLY!

for the horizontal direction, I'm not too sure?

@Fewmet- wouldn't gravity affect the horizontal speed because of the angles of the vectors?

 

[Fewmet]

@Fewmet- wouldn't gravity affect the horizontal speed because of the angles of the vectors?

No. I know that it seems, intuitively, like the downward pull of gravity makes something like a horizontally thrown ball speed up. What is increasing it the sum of the horizontal and vertical velocities.

Consider tossing a coin vertically while on a uniformly moving train. If gravity caused a horizontal acceleration, the coin would move forward faster than the train, than you and than your hand. It would land in front you on the train. The fact that it falls back into your hand is consistent with the principle that perpendicular vectors act independently of each other.

 

[rwisz]

I would first identify the given information and variables in the problem. The given information is the initial velocity (V = 500 m/s) and the launch angle (30º). The variables in the problem are the distance (D), time (t), and acceleration (a).

Next, I would use the equations of motion to solve for the unknown variables. Since we are given the initial velocity and angle, we can determine the horizontal and vertical components of the velocity using trigonometric functions. The horizontal velocity (Vx) can be calculated as Vx = Vcos(30º) and the vertical velocity (Vy) can be calculated as Vy = Vsin(30º).

Using the equation v = u + at, we can solve for the time (t) it takes for the projectile to reach its maximum height. Since the projectile is at its maximum height when its vertical velocity is equal to 0, we can set Vy = 0 and solve for t. This will give us the time it takes for the projectile to reach its maximum height.

To determine the maximum height achieved by the projectile, we can use the equation Vf^2 = Vo^2 + 2aD, where Vf is the final velocity (0 m/s), Vo is the initial velocity (500 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and D is the distance traveled in the vertical direction (maximum height).

To find the distance in the horizontal direction and the angle of the velocity vector when the projectile is -50m from the launch point, we can use the equations D = Vx*t and tanθ = Vy/Vx, where θ is the angle of the velocity vector. We can substitute the values we have calculated for Vx and t to solve for D, and then use the calculated values for Vx and Vy to solve for θ.

In conclusion, using the given initial velocity and angle, we can use the equations of motion to determine the time, maximum height, and distance and angle of the velocity vector when the projectile is -50m from the launch point. It is important to carefully identify and use the correct equations and variables to solve the problem accurately.