A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

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.