Question

solve this question ​

Answer 1

Answer:

option (c) : equal to each other

Explanation:

At any time t, a projectile's horizontal and vertical displacement are:

x = V•t•Cosθ where V is the initial velocity, θ is the launch angle

y = V•t•Sinθ – ½gt^2

The velocities are the time derivatives of displacement:

Vx = VCosθ (note that Vx does not depend on t, so Vx is constant)

Vy = VSinθ – gt

Velocity = Vxi + Vyj

The magnitude of velocity is √(Vx^2 + Vy^2)

At maximum height, Vy = 0 = VSinθ – gt

So at maximum height, t = (VSinθ)/g [total flight time = T = 2VSinθ/g]

The range R of a projectile launched at an angle θ with a velocity V is:

R = V^2 Sin2θ / g

The maximum height H is

H = V^2 Sin^2(θ) / 2g

You can see that the range of θ = 30° is a function of Sin60 and the range of θ = 60° is a function of Sin120

Sin60 = 0.866 and Sin120 = 0.866

So, yes, the range for θ = 30° and for θ = 60° is the same

Answer 2

Answer:

option c is correct.........