Question

Problem 11.104 A golfer hits a golf ball with an initial velocity of 160 ft/s at an angle of 25° with the horizontal. Knowing that the fairway slopes downward at an average angle of 5°, determine the distance d between the golfer and Point B where the ball first lands.

Answer 1

Answer:

Explanation:

Given

Initial launch velocity $v=160\ ft/s$

Launch angle $\theta =25^{\circ}$

slope angle $\phi =5^{\circ}$

Let d be the distance on the slope after which projectile lands

therefore horizontal distance covered by projectile $x=d\cos 5$

Vertical distance covered by projectile $y=d\sin 5$

Considering horizontal motion

Initial horizontal velocity $v_x=v\cos 25$

$d\cos5=160\cos 25\times t$

$d=145.56t---1$

considering vertical motion

Initial vertical velocity $v_y=v\sin 25$

$-d\sin 25=160\sin25\cdot t-\frac{1}{2}gt^2$

$-0.087t=67.60t-16.087t^2$

$67.62t-16.087t^2+0.087d-----2$

put value of t in equation 2

$80.28t-16.087t^2=0$

$t(80.28-16.087t)=0$

$t=4.99\s\ as\ t\neq 0$

substitute the value of t in equation 1

$d=145.56\times 4.99=726.4\ ft$