The launch angle (with the horizontal) in which the projectile can attain the longest range.?

Question

Please help me with my assigment!!!

Answer 1

The horizontal range is
R = u cos θ * T where T is the time of flight.
T = 2 u sin θ / g
Hence
R = u^2 * 2 sin θ cos θ / g
R = [u^2/g] sin 2θ
Since [u^2/g] is a constant, R depends only upon the value of sin 2θ
We know that the minimum value of sin 2θ is zero and the maximum value of sin 2θ is 1.
If 2θ varies from zero to 90 degree, sin 2θ varies from zero to 1.
If θ = 45°, 2 θ = 90° and sin 2 θ = sin 90 = 1.
If θ is changed from zero to 45°, R attains the maximum value of [u^2/g] at 45°,
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{For values more and less than 45, R is less than this value of [u^2/g].
For example if θ = 15, sin 2*15 = sin 30 = 0.5. The range is half the maximum value.
if θ = 75 sin 2*75 = sin 150 = 0.5. The range is again half the maximum value.
Hence for angles more and above 45 by equal amounts, the range is the same but is less than the maximum value of 45 degree}.

Answer 2

its an projectile situation.the ball will flow in a curved course..for this reason projectile action in 2d with the intention to succeed in the receiver ..the gap could properly be taken as selection of the projectile A) USE formulation R={u^2 sin(2?)]/g the place ? is preliminary perspective of 40 4 tiers to the floor R=31.4m locate u b) use H={u^2sin^2(?)2g} right here u is ordinary from above step attempt your self :)

Answer 3

♠ thus the projectile is launched with initial velocity
v = v1*i +v2*j, where |v|=1,
v1=cos(b) is horizontal component,
v2=sin(b) is vertical component;
♣ at time t vertical position of projectile is
y(t)=v2*t –0.5g*t^2;
it’ll hit the ground when y(t)=0 again,
hence t=2*v2/g is lapse of time to cover a distance
x=v1*t = v1*(2*v2/g) = 2*cosb*sinb/g =2*sin(2b) /g;
♦ max x should be when dx/db = 4*cos(2b)/g =0,
hence cos(2b)=0, 2b=90°, b=45°;