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.