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Find time of flight of projectile thrown horizontally with speed 10 metres per secon from a long inclined plane which makes an angle of = 45 the time of flight is : please explain with diagram
Please draw a square. A|'''''''''|B C """"" D Draw a straight line connecting A and D.The angle ADC is 45°. The paricle is thrown from point A in the direction AB(horizontally).So at t=0 or initially the particle has only horizontal component of velocity (say v)which is 10m/s and at t=0 the vertical component of velocity(say u) is zero. Lets say AC=h and after a time t the particle lands on ground.So t is time of flight. h=ut+(1/2)gt2 or,h=gt2/2=10t2/2=5t2 Now we have to calculate height h.Lets say CD=x. So tan45°=h/x or,1=h/x Or,h=x Now x=vt=10t So using equation h=x, 5t2=10t or,t=2s Please approve the answer if helped.
Time for flight is 2 u sin theta/gS for inclined plane t=2u sin theta/gT=2*10 sin 45/ g sin 45T=2sec
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