Q.
A projectile is thrown from a point O on the ground at an angle 45o from the vertical and with a speed 52m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity g =10 m/s2.The value of t is _______The value of x is _______
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answer is [OBJECT OBJECT], [OBJECT OBJECT].
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Detailed Solution
→from O to H time of ascent =Uyg=510=0.5sec→at highest point mvx=m2vx1vx1=10 m/s→ one part A of mass m/2 is falling vertically down in 0.5 sec⇒ part B is not having vertical component of velocity, it is falling freely in gravity⇒t=0.5secOQ=OP+PQ=5×0.5+10×0.5=7.5m→from O to H time of ascent =Uyg=510=0.5sec→at highest point mvx=m2vx1vx1=10 m/s→ one part A of mass m/2 is falling vertically down in 0.5 sec⇒ part B is not having vertical component of velocity, it is falling freely in gravity⇒t=0.5secOQ=OP+PQ=5×0.5+10×0.5=7.5m
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