Question

a projectile of mass 1 kg is projected with velocity of under root 20 m per second such that is right on the same level as the point of projection at the distance of root 3 which of the following options are correct the maximum height reached by the projectile can be 0.25 M the minimum velocity during its motion can under root 15 M per second the time taken for the flight can be under root 3 upon 5 second the maximum angle of projection can be 60 degree

Answer 1

The angle is such that, for that given speed of projection, exactly mid way between the two walls the maximum height must be achieved. We use this condition to find the value of the angle.

Suppose x is the distance from the wall from which the ball is thrown, then half the horizontal journey is x + h, at this point the ball must also reach the maximum height and it's vertical velocity must be zero.

0 = 2√(gh)sinA - gt

t = ( 2√(gh)sinA )/g … time for half journey

In this time, the particle covers distance x+h horizontally, hence

x + h = (2√(gh)cosA)*t … (dist = speed*time)

x + h = (2√(gh)cosA)*(2√(gh)sinA)/g

x + h = 2(gh)*(2cosAsinA)/g

x + h = 2hsin(2A)

x = h*(2sin(2A) - 1) …… { eqn 1 }

Let's find the time taken to cover only the horizontal distance x now. In this time the particle must have achieved the vertical height of H.

t' = x / (2√(gh)cosA) …. (time = dist / speed)

In this time particle is at height H, hence

H = 2√(gh)sinA*x/(2√(gh)cosA) - 0.5*g*t’^2

H = tanA*x - 0.5*g*x^2/(4gh(cosA)^2) … { eqn 2}

Now substitute eqn 1 in eqn 2 to eliminate x and solve the equation for the angle A.

I leave the rest (and the tedious) to you.. :p

Or may be I will solve as time permits and edit the answer later to include the final value of A.