1. ## projectiles?

A golfer hits a ball from point A with initial velocity 24.5 m/s at and angle $\alpha$ above the horizontal, where sin $\alpha$ = 0.8. The ball just clears the tops of the two trees. The tops of the trees are both 14.7m above the level of a and are horizontal distance dm apart.

a i) Find the time taken for the ball to reach the top of the first tree.

ii) Determine the value of d.

Ok i have worked this question out but i'm not confident the answer is right as i found it very hard. This is what i did.

v = 24.5
sin $\alpha$ = 0.8
tree h = 14.7

so using trig i get 14.7/hypotenuse = 0.8

hypotenuse = 14.7/0.8 = 18.475

cos $\alpha$ = adjacent/18.375

$18.375^2 = 14.7^2 + a^2$

a= $\sqrt{18.375^2 - 14.7^2}$

$\therefore a = 11.025$

$cos\alpha = \frac{11.025}{18.375}$

cos $\alpha$ = 0.6

x = $(vsin\alpha)t - 4.9t^2$

$(24.5*0.8)t - 4.9t^2 = 14.7$

$19.6t - 4.9t^2 = 14.7$

$t^2 - 4t + 3 =0$

$(t-3)(t-1)$

t = 3 or t=1

so t=1 at top of first tree.

time to second tree = 3

$(vcos\alpha)t = x_2$ where $x_2$ = distance to second tree

$24.5 * 0.6 * 3 = x_2$

$x_2 = 44.1$

$x_2 - x_1 = d$

$44.1 - 11.025 = 33.075$

could anybody confirm this is done right? or tell me where i have gone wrong if not?

2. Originally Posted by djmccabie
A golfer hits a ball from point A with initial velocity 24.5 m/s at and angle $\alpha$ above the horizontal, where sin $\alpha$ = 0.8. The ball just clears the tops of the two trees. The tops of the trees are both 14.7m above the level of a and are horizontal distance dm apart.

a i) Find the time taken for the ball to reach the top of the first tree.

ii) Determine the value of d.

Ok i have worked this question out but i'm not confident the answer is right as i found it very hard. This is what i did.

v = 24.5
sin $\alpha$ = 0.8
tree h = 14.7

so using trig i get 14.7/hypotenuse = 0.8

hypotenuse = 14.7/0.8 = 18.475

cos $\alpha$ = adjacent/18.375

$18.375^2 = 14.7^2 + a^2$

a= $\sqrt{18.375^2 - 14.7^2}$

$\therefore a = 11.025$

$cos\alpha = \frac{11.025}{18.375}$

cos $\alpha$ = 0.6
Uhh, $sin^2(\alpha)+ cos^2(\alpha)= 1$
$.8^2+ cos^(\alpha)= .64+ cos^2(\alpha)= 1$
$cos^(\alpha)= 1- .64= .36$
$cos(\alpha)= \pm .6$
and since $\alpha$ must be in the first quadrant, $cos(\alpha)= .6$.

x = $(vsin\alpha)t - 4.9t^2$
You are using x to represent height? Peculiar, but okay.

[tex] $(24.5*0.8)t - 4.9t^2 = 14.7$

$19.6t - 4.9t^2 = 14.7$

$t^2 - 4t + 3 =0$

$(t-3)(t-1)$

t = 3 or t=1

so t=1 at top of first tree.

time to second tree = 3

$(vcos\alpha)t = x_2$ where $x_2$ = distance to second tree

$24.5 * 0.6 * 3 = x_2$[/quote]
Okay, the horizontal speed is constant so this is correct

$x_2 = 44.1$

$x_2 - x_1 = d$

$44.1 - 11.025 = 33.075$

could anybody confirm this is done right? or tell me where i have gone wrong if not?
It looks good to me.

3. i got it right..? feel like i've finally made some progress!