• Best of luck to the class of 2025 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here

Projectile Help Asap (1 Viewer)

survivor

Member
Joined
Nov 10, 2002
Messages
110
help anyone i have these two questions due morrow and i cant do them

1.) a projectile is fired with initial velocity Vm/s at an angle of projectiion @, from a point 0 on horizontal ground. After 1sec, it just passes over a 12m high wall 20m form the point of projection.
a. find V and @
b. find the max height reached by the projectile
c. find the range R


2.)a shell is fired from the ground at an aeroplane flying at 400km/h at the instant when it is vertically overhead. The initial speed of the hsell is 1000m/s.
a. find the angle of projection @ of the shell, if it eventually hits its tagret.
b. if the aeroplane flies at a constant height of 2 km, find the time when it is hit by the shell..


tankyou to anyone who can help i just have no idea!!
 

PoLaRbEaR

The Bear
Joined
Apr 13, 2003
Messages
253
Location
Sydney
Gender
Male
HSC
2003
1)a)* x = Vtcos and y = -(gt)/2 + Vtsin

*at t = 1 and taking g = 10m/s...and x=20 and y=12 sub:
x = Vtcos => 20=Vcos
y = -(gt)/2 + Vtsin => 12= -5 +Vsin

*do simultaneous: => V=20/cos
12 = -5 + 20tan
tan = 17/20
= 40 22' [to nearest min.]

V = (20)/(cos40 22')
V = 26.25m/s [to 2dp]

b)*first find flight time, ie y=0
0=-5t + (26.25) t (sin40 22') [rearrange]
5t - 17t = 0
t=0 or t= 17/5 seconds
*time to reach max height is half flight time, ie 17/10 seconds
max height = -5(17/10) + (26.25) (17/10) (sin40 22')
= 289/20 m

c)*flight time equals 17/5
range = (26.25) (17/5) (cos40 22')
= 68m
 
Last edited:

OOOPPPs

New Member
Joined
Feb 6, 2003
Messages
7
2)a)

The aeroplanes position must equal to the projectiles position for it to collide.

therefore.. planes x position = projectiles x position

(edit: forgot to change the UNITS)

400 km / hr = 400 000 / 3600 = 111.11 m/s

hence: 111.11 t = 1000 t cos A
cancel out t
111.11/1000 = cos A
1/9 = cos A
A = 83.62

b)
h = 2000
projectiles y position = 1000 t sin A - 1/2 g t^2
sub in A = 83.62 and y = 2000
then use quadratic eqn

i dunno if its rite tho...
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top