A body is projected so that on its upward path it passes through a point x meters and y meters vertically from the point of projection. Show that if R is the range on the horizontal through point of projection the angle of elevation is arctan((y/x)(R/R-x)).
A body projected upward from the level ground at an angle of 50ยบ with the horizontal has an initial speed of 40 m/s. (a) How long will it take to hit the ground? (b) How far from the starting point will it strike? (c) At ehat angle with the horizontal will it strike?
To solve this problem, let's start by setting up a coordinate system. Assume that the point of projection is the origin (0,0) on the coordinate plane. We are given two points through which the projectile passes: (x, y) and (R, 0), where R is the range.
Now, let's consider the motion of the projectile. As it moves upwards, it reaches its highest point and starts to descend. At the highest point, its vertical velocity is zero. From the highest point, the projectile falls back to the ground.
We can analyze the motion of the projectile using basic principles of projectile motion. The horizontal and vertical components of the projectile's motion are independent of each other. The equations of motion for the horizontal and vertical directions are as follows:
Horizontal motion:
x = R * cos(theta) ---(1)
Vertical motion:
y = R * sin(theta) - (1/2) * g * t^2 ---(2)
Where:
- x represents the point through which the projectile passes horizontally.
- y represents the point through which the projectile passes vertically.
- R is the range on the horizontal through the point of projection.
- theta is the angle of projection (elevation).
- g is the acceleration due to gravity.
- t is the time of flight.
Let's rearrange equation (1) to solve for t:
t = x / (R * cos(theta))
Substitute this value of t into equation (2):
y = R * sin(theta) - (1/2) * g * (x / (R * cos(theta)))^2
Simplifying the equation:
y = R * sin(theta) - (1/2) * g * (x^2 / (R^2 * cos^2(theta)))
Multiply both sides by 2 * R^2 * cos^2(theta):
2 * R^2 * cos^2(theta) * y = 2 * R^3 * sin(theta) * cos^2(theta) - g * x^2
Divide both sides by 2 * R^2 * cos^2(theta):
y / (R * cos^2(theta)) = R * sin(theta) - (1/2) * g * (x^2 / R^2)
Using the trigonometric identity cos^2(theta) = 1 - sin^2(theta), we can substitute it in the equation:
y / (R * (1 - sin^2(theta))) = R * sin(theta) - (1/2) * g * (x^2 / R^2)
Rearrange the equation:
y / (1 - sin^2(theta)) = R * sin(theta) - (1/2) * g * (x^2 / R)
Since 1 - sin^2(theta) = cos^2(theta), we have:
y / cos^2(theta) = R * sin(theta) - (1/2) * g * (x^2 / R)
y = (R * sin(theta) - (1/2) * g * (x^2 / R)) * cos^2(theta)
Now, we simplify the equation to isolate the angle of elevation (theta):
Divide both sides by cos^2(theta):
y / cos^2(theta) = R * sin(theta) - (1/2) * g * (x^2 / R)
y = R * sin(theta) * cos^2(theta) - (1/2) * g * (x^2 / R) * cos^2(theta)
y = R * sin(theta) * cos^2(theta) - (1/2) * g * x^2 * cos^2(theta) / R
y = R * (sin(theta) - (1/2) * g * x^2 / R) * cos^2(theta)
Since tan(theta) = sin(theta) / cos(theta), we have:
y = R * (tan(theta) - (1/2) * g * x^2 / (R^3 / cos^2(theta)))) * cos^2(theta)
y = R * tan(theta) * cos^2(theta) - (1/2) * g * x^2 * cos^2(theta) / (R / cos^2(theta))
y = R * tan(theta) * cos^2(theta) - (1/2) * g * x^2 * cos^4(theta) / R
Divide both sides by R:
y / R = tan(theta) * cos^2(theta) - (1/2) * g * x^2 * cos^4(theta) / (R^2)
Finally, divide both sides by cos^2(theta):
y / (R * cos^2(theta)) = tan(theta) - (1/2) * g * x^2 * cos^2(theta) / R
Comparing this equation with the original equation involving arctan, we can see that:
arctan((y/x)(R/R-x)) = arctan(tan(theta) - (1/2) * g * x^2 * cos^2(theta) / R)
Therefore, we have successfully shown that the angle of elevation is arctan((y/x)(R/R-x)).