Question

A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball from the flat section it lays on, up a 45∘ ramp, over a moat, and into the hole which is ????=1.45 m away from the end of the ramp. If the opening of the hole and the end of the ramp are at the same height, y=0.640 m, at what speed must the golfer hit the ball to launch the ball over the moat so that it lands directly in the hole? Assume a frictionless surface, so the ball slides without rotating. The acceleration due to gravity is 9.81 m/s2

Answer 1

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$1.45=\frac{u^2\sin 90}{9.8}$

u=3.77 m/s

Conserving Energy

$E_{bottom}=E_{initial\ point\ at\ cliff}$

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

$\frac{mv^2}{2}=mgh+\frac{mu^2}{2}$

$v^2=u^2+2gh$

$v=\sqrt{u^2+2gh}$

$v=\sqrt{3.77^2+2\time 9.8\times 0.64}$

$v=\sqrt{26.75}=5.17 m/s$