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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
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Apr 13, 2003
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253
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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
 
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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...
 
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