A golfer gives a ball a maximum initial speed of 39.9 m/s.

[MrGoodyear812]

What is the longest possible hole-in-one for this golfer? Neglect any distance the ball might roll on the green and assume that the tee and the green are at the same level.

in m

and:

What is the minimum speed of the ball during this hole-in-one shot?

please explain your answers :D

thanks in advance

 

[xcvxcvvc]

What is the longest possible hole-in-one for this golfer? Neglect any distance the ball might roll on the green and assume that the tee and the green are at the same level.

in m

and:

What is the minimum speed of the ball during this hole-in-one shot?

please explain your answers :D

thanks in advance

You need to make two equations, one for height and one for x distance, based on some initial velocity and some angle. Can you think of what to do from there?

 

[Drew McFaul]

The maximum range that can be achieved will be done so when the ball is projected at an angle of 45 degrees to the horizontal. At this angle, both the horizontal and vertical velocities will be equal to 39.9*cos(45).

First, you should calculate the total time for which the ball is in the air.
Consider first the vertical motion of the ball. Neglecting air resistance and such, we can say that the time taken for the ball to reach its maximum height is equal to the time taken for the ball to fall back to the ground since the acceleration is constant (9.81m/s²).

The time to reach the maximum height can be calculated with the equation v=u + at where

v = 0 m/s
u=39.9*cos(45)
a=-9.81 m/s²
t= ?

Calculate t from this equation and double it to acquire the total time for which the ball is in the air.

Now consider the horizontal motion of the ball. The horizontal velocity is equal to 39.9 * cos(45). Using the time of flight which you just calculated, work out the horizontal distance travelled. (distance = speed * time)

This will be your answer since you do not have to worry about the ball bouncing or rolling.