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Grade: 12
        
Find time of flight of projectile thrown horizontally with speed 10 metres per secon  from a long inclined plane whichmakes an angle of = 45 the time of flight is  :
 please explain with diagram
10 months ago

Answers : (2)

Susmita
425 Points
							
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.
10 months ago
Navya khandelwal
29 Points
							
Time for flight is 2 u sin  theta/g
S for inclined plane t=2u sin theta/g
T=2*10 sin 45/ g sin 45
T=2sec
 
6 months ago
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