When the angle of projection is 35 the range of the projectile is R now if the angle of the projection is its range will remain same
Answers
Answer:
I'll do the calculation assuming no wind resistance and flat terrain and assuming the projectile is fired from ground level. (It should be noted that none of these assumptions is particularly reasonable for a real ballistics problem. For a real problem, the calculation is much harder.)
Call the initial velocity v. Assume the angle is x. The vertical component is v sin(x). The time it takes for gravity to decrease this velocity to zero is v sin(x)/g. The total time aloft is twice as much since it takes the same time to fall down as it took to go up. So the time aloft is 2vsin(x)/g. The total range is then given by the horizontal component of the velocity multiplied by this time:
R=(2vsin(x)/g)∗vcos(x)=v2sin(2x)g
We need to find a second value of x that gives the same answer as x=30 degrees. Hence, we need to find a second value of x such that sin(2x) = sin(60). Because sin(120)=sin(60), we see that 2x = 120 or x = 60.
So the answer is 60 degrees.
Interestingly, we can solve this problem quite generally.
We seek two different values of x that give the same range. The range only depends on x in the product of sin(x) and cos(x). But cos(x) = sin(90-x). So the range has the product of sin(x) and sin(90-x). So for any firing angle (between 0 and 90), the complement of the angle (i.e. 90 minus the angle) will yield the same range.
Answer:
60°
I HOPE THIS WILL HELP YOU