Chipping from the rough, a golfer sends the ball over a 3.00m high tree that is

Question

Chipping from the rough, a golfer sends the ball over a 3.00m high tree that is 14.0m away.

1. At what launch angle greater than 54.0 degrees does the golf ball just barely miss the top of the tree in front of the green? Assume the ball has an initial speed of 13.5 m/s, and that the tree is 3.00m high and is a horizontal distance of 14.0m from the launch point. (got the answer)

2. Where does the ball land in the case described in #1?

3. At what launch angle less than 54.0 degrees does the golf ball just barely miss the top of the tree in front of the green? ( got the answer)

4. Where does the ball land in the case described in #3?

Explanation / Answer

Q1.

let angle of launch be theta.

initial speed is 13.5 m/s

let at time t, it just reaches tree's height.

then horizontal distance covered=horizontal speed*time

=13.5*cos(theta)*t

==>13.5*cos(theta)*t=14

==>cos(theta)*t=1.037

=>t=1.037*sec(theta)....(1)

vertical displacement=3 m

==>13.5*sin(theta)*t-0.5*9.8*t^2=3

==>13.5*sin(theta)*1.037*sec(theta)-4.9*1.037^2*sec^2(theta)=3

==>14*tan(theta)-5.2693*(1+tan^2(theta))=3

==>-5.2693*tan^2(theta)+14*tan(theta)-8.2693=0

solving for tan(theta), we get

tan(theta)=1.77054 or tan(theta)=0.88636

then corresponding theta values are:

theta=60.542 degrees or theta=41.552 degrees

as given that value of launch angle is higher than 54 degrees, then theta=60.542 degrees

part 2:

the distance where the ball lands is given by

horizontal range=v^2*sin(2*theta)/g

=13.5^2*sin(2*60.542)/9.8=15.927 m

part 3:

smaller angle is 41.552 degrees as obtained in part 1.

part 4:

the distance where the ball lands is given by

horizontal range=v^2*sin(2*theta)/g

=13.5^2*sin(2*41.552)/9.8=18.462 m

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