What are the formulas for projectile on inclined plane?
Answers
Explanation:
During time of flight, the horizontal velocity u cos α remains constant. The greatest distance of the projectile from the inclined plane is u2sin2 (α-β)/2gcosβ . A Particle is projected with a velocity 39.2 m/sec at an angle of 30o to an inclined plane (inclined at an angle of 45o to the horizontal).
Explanation:
1: The coordinate “x” is along the incline – not in the horizontal direction; and the coordinate “y” is perpendicular to incline – not in the vertical direction.
2: Angle with the incline
From the figure, it is clear that the angle that the velocity of projection makes with x-axis (i.e. incline) is “θ – α”.
Projectile motion up an incline
The projection from lower level.
3: The point of return
The point of return is specified by the coordinate R,0 in the coordinate system, where “R” is the range along the incline.
4: Components of initial velocity
ux=ucos(θ−α)
uy=usin(θ−α)
5: The components of acceleration
In order to determine the components of acceleration in new coordinate directions, we need to know the angle between acceleration due to gravity and y-axis. We see that the direction of acceleration is perpendicular to the base of incline (i.e. horizontal) and y-axis is perpendicular to the incline.
Components of acceleration due to gravity
The acceleration due to gravity forms an angle with y-axis, which is equal to angle of incline.
Thus, the angle between acceleration due to gravity and y – axis is equal to the angle of incline i.e. “α”. Therefore, components of acceleration due to gravity are :
ax=−gsinα
ay=−gcosα
The negative signs precede the expression as two components are in the opposite directions to the positive directions of the coordinates.
6: Unlike in the normal case, the motion in x-direction i.e. along the incline is not uniform motion, but a decelerated motion. The velocity is in positive x-direction, whereas acceleration is in negative x-direction. As such, component of motion in x-direction is decelerated at a constant rate “gsin α”.
Time of flight
The time of flight (T) is obtained by analyzing motion in y-direction (which is no more vertical as in the normal case). The displacement in y-direction after the projectile has returned to the incline, however, is zero as in the normal case. Thus,
Projectile motion up an incline
The projection from lower level.
y=uyT+12ayT2=0
⇒usin(θ−α)T+12(−gcosα)T2=0
⇒T{usin(θ−α)+12(−gcosα)T}=0
Either,
T=0
or,
⇒T=2usin(θ−α)gcosα
The first value represents the initial time of projection. Hence, second expression gives us the time of flight as required. We should note here that the expression of time of flight is alike normal case in a significant manner.
In the generic form, we can express the formula of the time of flight as :
T=∣∣∣2uyay∣∣∣
In the normal case, uy=usinθ and ay=−g . Hence,
T=2usinθg
In the case of projection on incline plane, uy=usin(θ−α) and ay=−gcosα . Hence,
T=2usin(θ−α)gcosα
This comparison and understanding of generic form of the expression for time of flight helps us write the formula accurately in both cases.
Range of flight
First thing that we should note that we do not call “horizontal range” as the range on the incline is no more horizontal. Rather we simply refer the displacement along x-axis as “range”. We can find range of flight by considering motion in both “x” and “y” directions. Note also that we needed the same approach even in the normal case. Let “R” be the range of projectile motion.
The motion along x-axis is no more uniform, but decelerated. This is the major difference with respect to normal case.
x=uxT−12axT2
Substituting value of “T” as obtained before, we have :
R=ucos(θ−α)X2usin(θ−α)gcosα−gsinαX4u2sin2(θ−α)2g2cos2α
⇒R=u2gcos2α{2cos(θ−α)sin(θ−α)cosα−sinαX2sin2(θ−α)}
Using trigonometric relation, 2sin2(θ−α)=1−cos2(θ−α),
⇒R=u2gcos2α[sin2(θ−α)cosα−sinα{1−cos2(θ−α)}]
⇒R=u2gcos2α{sin2(θ−α)cosα−sinα+sinαcos2(θ−α)}
We use the trigonometric relation, sin(A+B)=sinAcosB+cosAsinB ,
⇒R=u2gcos2α{sin(2θ−2α+α)−sinα}
⇒R=u2gcos2α{sin(2θ−α)−sinα}
This is the expression for the range of projectile on an incline. We can see that this expression reduces to the one for the normal case, when α=0 ,
⇒R=u2sin2θg
Maximum range
The range of a projectile thrown up the incline is given as :
R=u2gcos2α{sin(2θ−α)−sinα}
We see here that the angle of incline is constant. The range, therefore, is maximum for maximum value of “sin(2θ – α)”. Thus, range is maximum for the angle of projection as measured from horizontal direction, when :
sin(2θ−α)=1
⇒sin(2θ−α)=sinπ/2
⇒2θ−α=π/2
⇒θ=π/2+α/2
The maximum range, therefore, is :
⇒Rmax=u2gcos2α(1−sinα)