In the given figure, particle A moves along the line y = 30 m with a constant velocity v→ of magnitude 3.0 m/s and parallel to the x-axis. At the instant particle A passes the y-axis, particle B leaves the origin with a zero initial speed and a constant acceleration a→ of magnitude 0.40 m/s2. What angle θ (in degrees) between a→ and the positive direction of the y-axis would result in a collision?

Collision between particles A and B requires two things.y=12ayt2⇒30m=120.40m/s2cos⁡θt2           .....(i)vt=12axt2⇒(3.0m/s)t=120.40m/s2sin⁡θt2We eliminate a factor of t in the last relationship and formally solve for time: t=2vax=2(3.0m/s)0.40m/s2sin⁡θThis is then plugged into Eq. (i) to produce30m=120.40m/s2cos⁡θ2(3.0m/s)0.40m/s2sin⁡θ2which, with the use of sin2⁡θ=1−cos2⁡θ, simplifies to 30=9.00.20cos⁡θ1−cos2⁡θ⇒1−cos2⁡θ9.0(0.20)(30)cos⁡θWe use the quadratic formula (choosing the positive root) to solve for cos θ:cos⁡θ=−1.5+1.52−4(1.0)(−1.0)2=12which yields θ=cos−1⁡12=60∘.