Projectile on an Incline question

[funzone36]

Question:

A projectile is fired down an incline which makes an angle of 40 degrees to the horizontal. The initial velocity of the projectile has a speed of 10m/s and makes an angle 60 degrees to the incline. Calculate the maximum perpendicular distance of the projectile from the incline.

Note:

I don't even know how to start solving this question and I'm not sure which formulas I should be using. If anyone can provide some hints as to how to start solving this question and the methods to get to the solution, that would be very helpful. Thanks.

 

[Kurdt]

The equations you use are just the normal kinematic equations. It is a bit more complicated with all the angles involved. Drawing a picture of this will also help you visualise the question better.

 

[baba89]

I would suggest starting by breaking down the problem into smaller, more manageable parts. First, we need to understand the motion of the projectile in this scenario. The projectile is being fired down an incline, meaning it has two components of velocity - one parallel to the incline and one perpendicular to the incline. The initial velocity of the projectile is 10m/s at an angle of 60 degrees to the incline. This means that the initial velocity can be broken down into two components: 10m/s * sin(60) = 8.66m/s parallel to the incline and 10m/s * cos(60) = 5m/s perpendicular to the incline.

Next, we can use the equations of motion to determine the trajectory of the projectile. Since the projectile is being launched at an angle, we will need to use the equations for projectile motion on a slope. These equations take into account the acceleration due to gravity and the angle of the incline.

The maximum perpendicular distance of the projectile from the incline will occur at the highest point of the projectile's trajectory. To find this point, we can use the equation:

y = y0 + v0y * t - 1/2 * g * t^2

Where y is the vertical position, y0 is the initial vertical position, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.

To find the time at which the projectile reaches its maximum height, we can use the equation:

vfy = v0y - g * t

Where vfy is the final vertical velocity, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.

Setting vfy to 0, we can solve for t and find that the time at which the projectile reaches its maximum height is 1.02 seconds.

Plugging this time into the first equation, we can find the maximum perpendicular distance of the projectile from the incline:

y = 0 + 5m/s * 1.02s - 1/2 * 9.8m/s^2 * (1.02s)^2 = 2.55m

Therefore, the maximum perpendicular distance of the projectile from the incline is 2.55m. I hope this helps you understand how to approach and solve this problem.