SHM and Projectile Motion (1 Viewer)

[Smithereens]

I've got some questions that I require assistance with. Thanks for any help

1. A projectile fired from a point on a horizontal plane reaches a maximum height 'h' before striking the ground 'd' metres away from its point of projection. Show that the equation of the path of the projectile may be expressed as:

y = 4hx (d-x) / d^2

2. It may be assumed that the tides rise and fall in SHM. A ship requires 10 m of water to safely enter a harbour. At low tide, the entrance is 9m deep and at high tide, 12m deep. In the first week of December, low tide is at 10:30am and high tide is at 5:30pm. During which times may the ship proceed safely?

[Ans: 1:14pm to 9:46pm]

Thanks.

 

[lyounamu]

I've got some questions that I require assistance with. Thanks for any help

1. A projectile fired from a point on a horizontal plane reaches a maximum height 'h' before striking the ground 'd' metres away from its point of projection. Show that the equation of the path of the projectile may be expressed as:

y = 4hx (d-x) / d^2

2. It may be assumed that the tides rise and fall in SHM. A ship requires 10 m of water to safely enter a harbour. At low tide, the entrance is 9m deep and at high tide, 12m deep. In the first week of December, low tide is at 10:30am and high tide is at 5:30pm. During which times may the ship proceed safely?

[Ans: 1:14pm to 9:46pm]

Thanks.

1.

y'' = -g
y' = -gt + vsin@
y = -1/2 . gt^2 + vtsin@

x'' = 0
x' = vcos@
x = vtcos@

t = x/vtcos@

Put that into y, you get:

y = xtan@ - gx^2/2v^2 . (1+tan^2(@)) ... (1)

When y = h, x = d/2 (because maximum height is at the middle)
h = d/2tan@ - g(d/2)^2/2v^2

When x=d, y =0
0 = dtan@ - gd^2/2v^2 . (1+tan^2(@))

Equate you get (I mean simultaneous equation):

h = tan@ (d/4)
so tan@ = 4h/d

Substitute that to the (1)

you get: y = 4h/d . x - gx^2/2v^2 ( 1+ 16h^2/d^2)
= 4hx/d - gx^2/2v^2 - 16gx^2h^2/2d^2v^2
= (4hxd - gx^2d^2/2v^2 - 16gx^2h^2/2v^2)/d^2
= 4hx(d-x)/d^2 since 4h = ((gd^2-16gh^2)/2v^2)

OP: Can you check if your solution is right by the way?


2)

Let 10:30 AM = 0 hour
And 5:30 PM = 7 hours

& Let 9 m = -1.5 and 12m = 1.5

So 10 m = -0.5

So you draw the cosine graph that is upside down.
The equation should be: h = -1.5cos(nt)

Check: When t=0, h=-1.5
-1.5 = -1.5cos 0 which is correct
Now when t=7, h=1.5
1.5 = -1.5 cos 7n
cos 7n = -1
7n = 180
n = 180/7

Now you want to find the time when -0.5 occurs.

-0.5 = -1.5cos180/7 t
cos180/7 . t = 1/3
180/7 . t = 70.5287793... or (360-70.5287793...)
t = 2.74278586... hours or 11.2572141... hours

2.74278586...hours = 2 hours and 44 minutes. 2 hours and 44 minutes after 10:30 am = 1:44 pm. Same thing for 11.2572141... hours.

 

[lyounamu]

Sorry for poorly set-out working out. I got really rusty in this topic. I can barely do a decent question.