Answer:
Golf ball will go a maximum of 270.36 meter.
Explanation:
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g [tex]m/s^2[/tex] and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile = 2u sin θ/g
Horizontal motion:
We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 [tex]m/s^2[/tex] and time taken = 2u sin θ /g
So range of projectile, [tex]R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}[/tex]
Now in the given problem
A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go
u = 51.5 m/s, for maximum range θ = 45⁰
So maximum distance reached = [tex]\frac{51.5^2sin(2*45)}{9.81}=270.36 meter[/tex]
So it will go a maximum of 270.36 meter.
