projectile question (1 Viewer)

[shkspeare]

A coastal defence cannon firse a shell horizontally from the top of a 50.0 m high cliff, directed out to sea with a velocity of 60 ms^-1.

Determine the range of the shell's trajectory

 

[Takuya]

191.7m easy

 

[shkspeare]

working out plz =)

 

[Inhuman]

You may want to show some working Takuya

 

[Takuya]

oh of course

Because it's horizontal, that's your horizontal velocity which doesn't change. So how do you get range? x velocity mutiplied by time.

The time in this case is the time it takes for the projectile to fall 50m to the ocean. i.e. use r=ut+0.5(at^2)

After that it's just substitution. I could do it mathematically for you as well, but this is Physics not Maths

 

[helper]

Calculate time of flight from vertical.

y=ut + 0.5*9.8*t^2
50 = 0 + 4.9t^2

t= 3.19

Range from horizontal.

x=uxt
= 3.19*60
=191.7m

 

[Takuya]

yep that's it.

 

[shkspeare]

thx guys

 

[shkspeare]

ok how about if the same cannon (50m high) was fired at an angle of 45 degrees at 60m/s wot is the new range

 

[walla]

i get 411.9m

 

[walla]

i use
y = ut + 0.5at^2
y = 50 (take down to be positive)
u = -60sin45
= -60/sqrt2
a = 9.8
use quadratic formula to find t
substitute into x = ut
u = 60cos45
= 60/sqrt2
and you get 411.9
did that pretty quickly so there could be a mistake somewhere but the answer seems about right

 

[helper]

Calculate inital x and y velocities.

Solve same way with new values of u

Solve as quadratic for t

411.6 is what I ended up with

 

[walla]

ooo 0.3 who's it gonna be

 

[helper]

Rounding. The answer they would want is 412. If they were being picky, they could take off a mark from both of us.

 

[shkspeare]

thx guys

how many d.p's should we use wen answering numerical questions

 

[helper]

significant figures same as question.
You quoted height as 50.0, so 3 sig figs