Question

From points A and B at the respective heights of 2m and 6m, two bodies are thrown simultaneously towards each other; one is thrown horizontally with a velocity of 8m/s and the other, downward at an
angle of 45o to the horizontal at an initial velocity such that the bodies 2m collide in flight. The horizontal distance between points A and B
equals 6m. The initial velocity v0 of the body thrown at an angle 45 theta is –

(A) 4 root 2m/s (B) 8 root 2m/s (C) 16 root 2m/s (D) 32 root 2m/s

Answer 1

Answer:

Explanation:

tanθ=48⇒tanθ=12…..(1)

v→A=8iˆ,vB=−v→0cos45∘iˆ−v→0sin450jˆ

v→BA=(−v02–√−8)i−v02–√j

Direction of v→BA

tanθ=v02–√2–√(v0+82–√)....(2)

From eq (1) and eq (2) 2v0=v0+82–√,v0=11.28m/s

v→BA=(−v02–√−8)i−v02–√j

∵v0=82–√Ϸv→BA=−16iˆ−8jˆ

∣∣v→BA∣∣=t=SBA

((−16)2+(−8)2−−−−−−−−−−−−−−√)t=82+42−−−−−−√

t=80320−−−−√=14−−√⇒t=0.5s

c)x=vxt=(8)(0.5)=4

y'=12gt2=12×10×14=1.25

y=2−y'=0.75