Mechanics projectile M2 question
The trajectory of the particle is a parabola because of gravity. The acceleration due to gravity is 9.81m/s^2 and is, of course, a negative influence on the particle's motion.
If we draw a graph of the parabola and make it symmetrical, the point A is on the y axis at y=8, representing 8m. On the x axis at O(-10,0) and P(10,0) the parabola intersects the x axis. The point A(0,8) is the vertex at which the slope is horizontal. Let y=ax^2+bx+c. A is the y intercept so c=8. When y=0, x=10 and -10, so 0=100a+10b+8=100a-10b+8, making b=0, and 100a=-8, so a=-0.08, making the equation of the parabola y=8-0.08x^2.
We can relate x and t and y and t, where t is time. The general equation is s=ut-(1/2)gt^2, where s is distance, u an initial speed, and g is the acceleration of gravity which in this case acts against the particle's movement because it is initially projected against gravity. This is a scalar equation, but we need to apply it in a vector situation.
Now that we have the equation of the parabola, we need to find out the gradient at O(-10,0) to give us the projection angle. So differentiate y=8-0.08x^2 and we get y'=-0.16x. At x=-10 y'=1.6. The angle of projection z is given by tanz=1.6.
Velocity is a vector and the initial projection velocity can be split into a horizontal and vertical component. Gravity acts only on the vertical component. We can apply the motion equation by replacing s with y and u with vsinz, where v is the magnitude of the projection velocity. Therefore y=v(sinz)t-(1/2)gt^2. For x the motion equation is simply x=v(cosz)t.
We can use t to represent the time taken to go from O(-10,0) to P(10,0). The vertical displacement (y) then becomes 0, because the particle starts on the ground and ends on the ground. The horizontal displacement is 20m. By eliminating t between equations we can find the magnitude of v.
Putting y=0 in the motion equation: v(sinz)t=(1/2)gt^2, so t=2v(sinz)/g and x=v(cosz)t=2v^2(sinz)(cosz)/g.
The horizontal displacement (x) is 20m so 2v^2(sinz)(cosz)/g=20 and v=sqrt(10g/(sinzcosz)). We can calculate sinz and cosz from tanz. sinz=1.6/sqrt(3.56) and cosz=1/sqrt(3.56), so sinzcosz=1.6/3.56. Therefore v=sqrt(98.1*3.56/1.6)=14.774m/s. z, the projection angle=tan^-1(1.6)=58 degrees.
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