Question

A particle A is projected with an initial velocity of 60 m/s at an angle 30∘ to the horizontal. At the same time, a second particle B is projected in direction as shown in figure with initial speed of 50 m/s from a point at a distance of 100 m from A. If the particles collide in air, find the time when the particles collide.

A
1.09 s
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B
1.38 s
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C
1.53 s
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D
1.72 s
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Solution

The correct option is A 1.09 s As both the particles collide in the air, so Vertical height covered by the particle A in time t= Vertical height covered by the particle B in time t ⇒hA=hB ⇒uAyt−12gt2=uByt−12gt2 ⇒uAy=uBy ⇒uAsin30∘=uBsinα ⇒sinα=35 or cosα=45 Taking x and y directions as shown in figure |uAB|=uAx−uBx=60cos30∘−(−50cosα) =(30√3+50×45) m/s =(30√3+40) m/s Therefore, time of collision is t=AB|uAB|=10030√3+40 or t=1.09 s

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