3rd type of projectile question (1 Viewer)

[wolf7]

i got trouble with 2 question

with the speed must a water skier leave a 15 degree ski ramp whose end is 1.5m above water level, if he is to land 10m beyond the end of the ramp

also the next question is

the hole on a golf course is 45m horizontally from the tee. if a golfer can hit the ball at 30m/s, at what angle must he hit the ball to get a hole in one

 

[jono_z1]

the hole on a golf course is 45m horizontally from the tee. if a golfer can hit the ball at 30m/s, at what angle must he hit the ball to get a hole in one

I have no idea about question 1 either.

I got an answer for question 2, although it involves simultaneous equations and a trigonemetric property i just learnt about in 3 unit maths (*), so there must be an easier way.

(Let # be the angle we have to find)

u = 30m/s
u(horizontal) = 30cos#
u(vertical) = 30sin#

we can get an expression for time:
v = u + at
0 = 30sin# -9.8t (taking upwards as positive)
9.8t = 30sin#
t = (30sin#)/9.8 (this is for the way up only)

hence total time is 2t:
t = (60sin#)/9.8 (1)

range = u(horizontal) x t(vertical)
45 = 30cos# x t
t = 45/(30cos#)
t = 3/(2cos#) (2)

Solving simultaneously (1) and (2):
(60sin#)/9.8 = 3/(2cos#)
120sin#cos# = 29.4
sin#cos# = 0.245
2sin#cos# = 0.49

We know 2sin#cos# = sin(2#) (* 3 unit maths trig property)

hence sin 2# = 0.49
2# = sin^-1 0.49
# = 14.67°

There's a good java applet at http://www.walter-fendt.de/ph11e/projectile.htm which you can use to check your answers to some of these projectile motion questions.

 

[Riviet]

I tried Q2 as well, you could probably get 3 equations with 3 unknown and solve that but that would take much longer than solving 2 trig equations simultaneously which aren't hard in this question. They're not gonna ask one with so few information in the hsc. They will give you a minimum of 2 useful info. Question 1 i tried to get started but with no success, it probably requires some advanced maths that I haven't learnt at school yet so don't worry about the first question.

 

[Lalli]

I think this is just a tad late
i was struggling with Q1 as well and this is what i got!
Q1)
final displacement= 0
Uy = usin 15'

therefore using:
r = ut + 0.5at^2

0= usin15t -4.9t^2

t = usin15/ 4.9

Now using Uxt = delta x (horzontal component which we know is 10m)

Ux = ucos15

delta x= 10

ucos15t=10

t =10/ ucos15

Now simply substitute these equations

10/ucos15 = usin15/ 4.9

u^2 cos15sin15 = 49

u= sqrt (49/cos15sin15)

u= 14 ms-1

I WAS SOO ECSTATIC WHEN I WORKED IT OUT!! just wanted to share