Respuesta :
Answer:
The required angle is (90-25)° = 65°
Explanation:
The given motion is an example of projectile motion.
Let 'v' be the initial velocity and '∅' be the angle of projection.
Let 't' be the time taken for complete motion.
Let 'g' be the acceleration due to gravity
Taking components of velocity in horizontal(x) and vertical(y) direction.
[tex]v_{x}[/tex] = v cos(∅)
[tex]v_{y}[/tex] = v sin(∅)
We know that for a projectile motion,
t =[tex]\frac{2vsin(∅)}{g}[/tex]
Since there is no force acting on the golf ball in horizonal direction.
Total distance(d) covered in horizontal direction is -
d = [tex]v_{x}[/tex]×t = vcos(∅)×[tex]\frac{2vsin(∅)}{g}[/tex] = [tex]\frac{v^{2}sin(2∅) }{g}[/tex].
If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -
α +β = 90° as-
d = [tex]\frac{v^{2}sin(2α) }{g}[/tex] = [tex]\frac{v^{2}sin(2[90-β]) }{g}[/tex] =[tex]\frac{v^{2}sin(180-2β) }{g}[/tex] = [tex]\frac{v^{2}sin(2β) }{g}[/tex] .
∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.
∴ If α = 25° , then
β = 90-25 = 65°
∴ The required angle is 65°.
Answer:
- 25° from positive X axis .
Explanation:
If a projectile projects with velocity u and at angle of projection Ф, it strikes the ground with the same velocity and at same angle but in downward direction.
So, it makes an angle of - 25°.
