Guide for non-golfers: the tee is where the bal is launched from, and the green is where the golfer wants the ball to land. the green has a hole in it, and the golfer is trying to get the ball to go into the hole. the golfer inbee park is at the tee at the 16th hole, a short par 3. the tee is 18.0 m vertically above the level of the green, and some distance horizontally from the hole park grabs her 5-iron, and strikes the ball so that it is launched from the tee with an initial velocity of 42.0 m/s at an angle of 35.0° above the horizontal. the ball lands on the green at a distance of 5.00 m from the hole, and then, still moving away from the tee, bounces twice, and then rolls right into the hole. the crowd goes crazy-it's a hole in one! neglect air resistance, and assume that the acceleration due to gravity is 9.80 m/s2 (a) how long does the ball spend in the air? (the time it leaves the tee until it first lands on the green.) assume the green is perfectly horizontal. 1.917xs (b) what is the horizontal distance from the tee to the hole? 70.95xm

a) The ball is 5.58 s in the air.

b)  The horizontal distance from the tee to the hole is 197 m.

Explanation:

The vector that describes the position of the ball at time t can be calculated as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

x0: initial horizontal position

v0: initial velocity

t: time

α:  launching angle

y0: initial vertical position

g: acceleration due to gravity  (-9.80 m/s² considering the upward direction as positive)

Please, see the figure for a description of the situation. Notice that the frame of reference is located at the throwing point.

a) We know that at final time the vertical position of the ball is -18 m relative to the launching point (see figure). Then, using the equation for the y-component of the vector r, we can obtain the time of flight:

y = y0 + v0 · t · sin α + 1/2 · g · t²

-18.0 m = 0 m + 42.0 m/s · t · sin 35° - 1/2 · 9.8 m/s² · t²

0 = -4.9 · m/s² · t² + 42.0 m/s · t · sin 35° + 18.0 m

Solving the quadratic equation:

t = 5.58 s  

b) The horizontal distance traveled by the ball is given by the x-component (rx in the figure) of the vector r at final time. Then:

x = x0 + v0 · t · cos α

x = 0 m + 42.0m/s · 5.58 s · cos 35° = 192 m

Since the ball lands at 5 m of the hole, the horizontal distance from the tee to the hole is (192 m + 5.00 m) 197 m.

asthenosphere and convection

i think it would be 5