A particle P is launched from a point O with an initial velocity of 60m/s at an angle of 30 degrees to the horizontal. At the same time, a second particle Q is projected in the opposite direction with initial speed of 50m/s from a point level with point O and 100m from O. You are requested to determine the angle of projection of particle Q needed if the particles collide. Also determine when the collision occurs.
Using the usual connotation, we can say that: y = usin(θ)t-gt^2/2 = vsin(Φ)t-gt^2/2 or usin(θ)=vsin(Φ). Use the given values to obtain: 60*1/2 =50sin(Φ). In other words, sin(Φ)= 3/5.
The second projectile must therefore be launched at angle Φ~=36.87° to the horizontal.
Next,assume that the two projectiles meet after t secs. Then:t(ucos(θ))+t(vcos(Φ))= 100 or t(60√3/2 + 50*4/5)=100 which yields t~ =1.0874 sec
To solve this problem, we need to find the time and angle of projection for particle Q such that it intersects the path of particle P.
Let's start by analyzing the motion of particle P. From the given information, we know that its initial velocity is 60 m/s at an angle of 30 degrees to the horizontal. We can break this initial velocity into its horizontal and vertical components.
Horizontal component of velocity (Vpx) = Initial velocity * cos(angle)
Vpx = 60 m/s * cos(30 degrees)
Vpx = 60 m/s * √3/2
Vpx = 30√3 m/s
Vertical component of velocity (Vpy) = Initial velocity * sin(angle)
Vpy = 60 m/s * sin(30 degrees)
Vpy = 60 m/s * 1/2
Vpy = 30 m/s
Now, let's analyze the horizontal motion of particle P. Since there is no horizontal force acting on it, the horizontal velocity remains constant throughout its trajectory.
The horizontal distance covered by particle P (x) can be calculated using the formula:
x = Vpx * t
where t is the time taken for particle P to reach the collision point.
Now, let's analyze the vertical motion of particle P. The vertical motion can be described using the equation of motion:
y = Vpy * t - (1/2) * g * t^2
where y is the vertical distance covered by particle P, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Since the two particles collide, we know that the vertical positions of the two particles will be the same when the collision occurs. Therefore, we can set up an equation for the vertical positions of particles P and Q.
y(P) = y(Q)
Vpy * t - (1/2) * g * t^2 = -(1/2) * g * t_Q^2
where t_Q is the time taken for particle Q to reach the collision point.
Now, let's analyze the horizontal motion of particle Q. Since it is projected in the opposite direction with an initial velocity of 50 m/s, we can analyze its horizontal motion similarly to particle P.
Horizontal component of velocity for particle Q (Vqx) = Initial velocity * cos(angle_Q)
where angle_Q is the angle of projection of particle Q.
The horizontal distance covered by particle Q (x_Q) can be calculated using the formula:
x_Q = Vqx * t_Q
To determine the angle of projection angle_Q, we need to find the value of t_Q for which particle Q intersects with particle P. To simplify the calculation, we can substitute the expression for x_Q into the previous equation for x, and solve it for t_Q.
x = Vpx * t
x_Q = Vqx * t_Q
Since particle Q is projected from a point 100m away from O, the horizontal distance covered by particle Q (x_Q) can be expressed as:
x + x_Q = 100m
Substituting the values in the equation:
(Vpx * t) + (Vqx * t_Q) = 100m
Now, we have two simultaneous equations that we can solve: one for the vertical position and one for the horizontal position.
Vpy * t - (1/2) * g * t^2 = -(1/2) * g * t_Q^2 (Equation 1)
(Vpx * t) + (Vqx * t_Q) = 100m (Equation 2)
By solving these equations simultaneously, we can determine the values of t and t_Q. Once we have the values of t and t_Q, we can find the angle of projection (angle_Q) using the same technique as for particle P.
Finally, we can determine when the collision occurs by using the value of t (or t_Q) and substituting it back into the equation x = Vpx * t (or x_Q = Vqx * t_Q). This will give us the time at which the collision occurs.
Note: The calculations required to solve this problem may involve substitutions, rearrangements, and solving simultaneous equations, which can be quite involved.