Projectile Q (1 Viewer)

[vds700]

Consider an old military cannonsitting on a 70m cliff and aiming downwards at 15 degrees below the horizontal. A shot cannon ball hits the water below. Initial velocity of the ball is 85 m/s, find the time taken for the cannonball to hit the water?

I tried using v^2 = u^2 + 2as, but get a square root of a negative. In the soultion they just state that the final velocity is -43 and use v = u + at. I dont understand where they got it from.

Can someone please help. Thanks

PS Im just considering the vertical components of the motion

 

[3unitz]

-70 = t (85sin 15) - t^2 (9.8 / 2)

t^2 (9.8 / 2) - t (85sin 15) - 70 = 0

t ~ 6.6 sec

 

[Pwnage101]

Got it

ok

use delta(Y) = ut + 0.5*a*t^2

now, i'm gonna set it up as follows:

delta(Y) = -70
u=-85sin(15)
a=-9.8

sub in:

-70 = -85sin(15)t -0.5*9.8*t^2

re arrange

4.9t^2 + 85sin(15)t - 70 = 0

treat as a quadratic in t

(to avoid gettin messy, im gonna take 85sin(15) = 22)

use quadtratic formula -

t = [-22 +/- sqrt(484 - 4*4.9*-70) ]/9.8
= [-22 +/- sqrt(1856) ]/9.8
= [-22 +/- 43 ]/9.8

but, t>0 so can only be:

= [-22 + 43 ]/9.8
= 21/9.8
= 2.14 seconds

Now u can find out the velocity by using v = u + at

a piece of advice: to find the rigth equation to use, it must contain variables all of which u knwo expet one - u cant find time using v=u+at in thsi one because we dont know v, adn we're trying to find t (usually, if teh projectile is fierd up, time half is when v = 0)

so just to confirm, v=u+at

v = -22-9.8*2.14
= -22 - 20.98
= -43 (nearest figure) = 43 m/s downwards

 

[Pwnage101]

-70 = t (85sin 15) - t^2 (9.8 / 2)

t^2 (9.8 / 2) - t (85sin 15) - 70 = 0

t ~ 6.6 sec

uh, 3unitz, shouldnt it be ' -85sin15 ' in the first line????

 

[vds700]

Thanks guys. I realise my mistake- I put delta y = 70, not -70, and got a square root of a negative.

 

[vds700]

BTW is it acceptable to use the quadratic formula to solve a projectile motion question in a physics exam?