A projectile is shot at a hill, the base of which is 300 m...

A projectile is shot at a hill, the base of which is 300 m away. The projectile is shot at $60^{\circ}$ above the horizontal with an initial speed of $75 \mathrm{m} / \mathrm{s}$. The hill can be approximated by a plane sloped at $20^{\circ}$ to the horizontal. Relative to the coordinate system shown in the following figure, the equation of this straight line is $y=\left(\tan 20^{\circ}\right) x-109 . \quad$ Where on the hill does the projectile land?

Key Concepts

Projectile Motion

Projectile motion is the two-dimensional motion of an object that is launched into the air and influenced only by the force of gravity (assuming air resistance is negligible). This concept involves decomposing the motion into horizontal and vertical components, where the horizontal motion occurs at a constant velocity and the vertical motion is uniformly accelerated under gravity. It is fundamental in predicting the trajectory of any object thrown or launched into the air.

Kinematic Equations

Kinematic equations are mathematical formulas that describe the motion of objects under constant acceleration. In the context of projectile motion, these equations relate displacement, velocity, acceleration, and time for both the horizontal and vertical components of the trajectory. They are essential for calculating key parameters such as maximum height, range, and the time of flight of the projectile.

Equation of a Straight Line

The equation of a straight line in a two-dimensional coordinate system typically takes the form y = mx + b, where m represents the slope and b represents the y-intercept. This concept is used to describe linear relationships and is crucial in this context for representing physical surfaces, such as a hill, as inclined planes. Understanding this allows the determination of where a projectile's path (a parabolic trajectory) intersects the linear surface.

Intersection of Curves

The intersection of curves involves finding the point or points at which two different mathematical expressions share the same coordinates. This method is applied by setting the equation of one curve equal to the equation of the other, thereby solving for the unknown variables. In problems involving projectile motion intersecting with other surfaces like a hill, it is a key tool for determining impact or landing points.

Transcript

00:01 Here in this question we are going to calculate the distance at which the projectile hits the plane of the hill.

00:10 So now here we break the motion of the projectile into two components with a of x as zero and a of y as minus g.

00:21 Now on taking the horizontal distance covered by the projectile as x meter from origin and we take the final vertical position of the projectile as y meter from the ground level.

00:38 So by using the equation of motion that is y equals to y naught plus v of y naught multiplied by t plus 1 divided by 2 a of y multiplied by t square.

01:03 Here v of y naught is the initial vertical component of the velocity of projectile y naught is the initial vertical component of position of the projectile y is the final vertical composition component of position of projectile and a of y is the acceleration due to gravity.

01:21 So here we substitute 75 multiplied by sine 60 degree meter per second for v of y naught 0 for y naught and minus 9 .81 meter per second square for a of y.

01:42 So now here on substituting the value we get y equals to 0 meter plus 75 multiplied by sine 60 degree multiplied by t plus 1 divided by 2 multiplied by minus 9 .81 meter per second square multiplied by t square.

02:13 So now on simplifying we get y equals to 64 .95 multiplied by t minus 4 .905 multiplied by t square.

02:35 Now here we substitute tan 20 degree multiplied by x minus 1094 y.

02:42 Therefore we get tan 20 degree multiplied by x minus 109 equals to 64 .95 multiplied by t minus 4 .905 multiplied by t square.

03:05 Now on simplifying we get 0 .36 multiplied by x minus 109 equals to 64 .95 multiplied by t minus 4 .905 multiplied by t square.

03:32 Now here we use the equation of motion that is x equals to x naught plus v of x naught multiplied by t.

03:48 Here x is the final horizontal position of the projectile x naught is the initial horizontal position of the projectile and v of x naught is the initial horizontal component of velocity...

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