Question

a body is projected upward making an angle thita with horizontal with a velocity of 300 ms^-1 find the value of thita so that the horizontal range Will be maximum hence find it's range and time of flight​

Answer 1

Answer:

Maximum Range: 9000m

Time of flight: 42.4 seconds (approx)

Explanation:

Vertical component of velocity will be 300sinx

horizontal component of velocity will be 300cosx.

hence Range will be 300cosx

On applying y=ut+1/2at^(2) for y direction, we get T=60sinx

putting this value of T in range, we get

Range= 9000*2*sinx*cosx

...putting sin2x at the place of 2*sinx*cosx

Range = 9000 sin2x

for Range to be maximum, sin2x must be max

i.e sin2x = 1, this gives us x=45°

hence theta should be 45° for maximum range.

On putting this value of theta in Range, we get

Max Range = 9000 sin2*45°

Max Range = 9000 sin90°

Max Range = 9000 metres or 9 km

On putting this same value of theta in time, we get

Time = 60 sin45°

Time = 60/√2

Time = 42.43 seconds