at what angle does an object thrown will travel max distance?

Answer 1

Here are some details. object is thrown with an initial velocity v at an angle of T relative to the ground.

vsinT = vertical component of velocity
The time to reach max altitude is when the vertical component of the velocity is equalt to that provided by the acceleration of gravity. So:
vsin(T) = gt
t = v sin(T)/g

The total time of flight is 2t and the distance is this time times the horizontal velocity component which is vcos(T). So:
d = v cos(T) * 2t = vcos(T) vsin(T) /g
d = ( 2v^2/g) sin(T) cos(T)

dd/dT = 0 = (2v^2/g)(cos^2T - sin^2T) = 0
cos^2T = sin^2T = 1 - cos^2T
2 cos^2T = 1
cosT = sqrt(2)/2
So T = 45 degrees

Answer 2

The angle will be 45 degree

Answer 3

Range is the product of horizontal velocity x time of flight.

R = U cos θ *T


Time of flight is given by = 2 U sin θ/ g.

Hence R = 2 sin θ*cos θ U ^2 /g =[ sin 2 θ ]. U^2 /g

For a given initial speed R depends on the value of sin 2 θ
.

The maximum value for sin term is 1.

The maximum value of R = U^2 /g and this is when sin 2 θ =1.

In other words when 2 θ = 90 °

Or when θ = 45°

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The ranges will be less than the range for 45degree

The ranges will be the same for any angle greater or less than 45 by a given value
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Answer 4

To get maximum range the body should be projected at angle 45 degrees

Answer 5

45 degrees if you assume that air resistance is negligible. When air resistance is not negligible, the best angle would be a bit lower than 45 degrees.

Answer 6

45 degrees

Answer 7

45 degrees

Answer 8

45 degrees

Answer 9

I agree. I would think in all cases the answer would be 45 degrees.

Answer 10

45 degrees.