- #1
abdo99
- 2
- 0
Attempt solution
Y=(vsinø)t-0.5gt^2
4=(vsinø)t-4,9t^2
But there is 3 variables!
Another eq is
X=(vcosø)t
6=(vcosø)t
But i did not found any away to solution
Question in dynamics projectile motion
- Thread starter abdo99
- Start date
In summary, projectile motion is the motion of an object that is launched or thrown into the air and differs from regular motion due to its involvement of both horizontal and vertical components. The trajectory of a projectile can be calculated using equations of motion and the maximum height reached can be determined by finding the apex. The range of a projectile can be calculated using the initial velocity and launch angle. Air resistance can affect projectile motion by altering the object's velocities and trajectory, with a greater impact on objects with larger surface areas.
Attempt solution
Y=(vsinø)t-0.5gt^2
4=(vsinø)t-4,9t^2
But there is 3 variables!
Another eq is
X=(vcosø)t
6=(vcosø)t
But i did not found any away to solution
- #2
dauto
- 1,948
- 201
You posted a sideways picture
- #3
rtsswmdktbmhw
- 38
- 2
Hint: 1. The ball travels on a parabolic path.
2. If you know the max height of the ball, you can calculate the vertical and horizontal components of initial velocity.
Or, at least, that's how I managed to do it but there may be a simpler method/formula..
2. If you know the max height of the ball, you can calculate the vertical and horizontal components of initial velocity.
Or, at least, that's how I managed to do it but there may be a simpler method/formula..
Last edited:
- #4
abdo99
- 2
- 0
I know that ,but i did not konw the maximum H
The rule is H=(u^2sin^2)/2g
The rule is H=(u^2sin^2)/2g
- #5
HalfLostAndSearching
- 1,008
- 0
In order to solve this problem, we need to use the kinematic equations for projectile motion. These equations relate the displacement (x and y), initial velocity (v), initial angle (ø), time (t), and acceleration due to gravity (g). In this case, we have two equations for y and x, but we need to find the values for v, ø, and t.
To do this, we can use a system of equations, where we set the two equations equal to each other and solve for v and t. This will give us the initial velocity and time at which the projectile reaches a height of 4 meters and a horizontal distance of 6 meters.
Once we have the values for v and t, we can then use those values to solve for ø using the equation ø = tan^-1 (y/x). This will give us the initial angle at which the projectile was launched.
However, it's important to note that in order to fully solve this problem, we also need to know the initial position of the projectile and the angle at which it was launched. Without this information, we cannot fully determine the solution.
FAQ: Question in dynamics projectile motion
1. What is projectile motion and how does it differ from regular motion?
Projectile motion is the motion of an object that is launched or thrown into the air, such as a ball or a bullet. It differs from regular motion because it involves both horizontal and vertical components of motion, whereas regular motion only involves motion in one direction.
2. How is the trajectory of a projectile calculated?
The trajectory of a projectile can be calculated using the equations of motion, which take into account the initial velocity, acceleration due to gravity, and the time elapsed. These equations can be solved for the horizontal and vertical components of motion to determine the overall trajectory.
3. What is the maximum height reached by a projectile?
The maximum height reached by a projectile can be determined by finding the point where the vertical velocity component becomes zero. This occurs at the highest point of the trajectory, known as the apex.
4. What is the range of a projectile?
The range of a projectile is the horizontal distance it travels before it hits the ground. This can be calculated using the initial velocity and the angle of launch.
5. How does air resistance affect projectile motion?
Air resistance can affect projectile motion by slowing down the object's horizontal and vertical velocities, and therefore changing its trajectory. This is more significant for objects with larger surface areas, such as a feather compared to a bullet.