help wif a ques plz (1 Viewer)

[jimmik]

a projecticle is launched, with angle of elevation @, from an origin O on level ground. At t seconds, its horizontal and vertical displacement, x and y metres respectively, are given by:
x = 20tcos@, y = -5t^2 + 20tsin@

a) show by eliminating @, that x^2 + (y +5t^2)^2 = 400t^2
b) hence or otherwise find the maximum horizontal displacement that the projectile can attain 1 second afta being launched.
c) find the maximum vertical displacement that th eprojectile can reach1 second afta being launched

i can figure out part a) but how do u use it to answer parts b) and c)? id be grateful for any help, thnx

 

[Xayma]

Well you could do it as a maxima minima problem.

for c) it is obvious that the maximum vertical displacement would be launching it straight up as it is completley independent of x (whereas x stops when y=0).

But for b) I think that its makes distance will be when it hits the ground after 1 second. So you could work out the angle from that then stick it in the first formula.

Thats all I can suggest without learning projectile motion early.

 

[CM_Tutor]

Put t = 1 into the equation derived in (a). You get x² + (y + 5)² = 400.

Forget about motion for a second, and think about this equation. It is a circle.

Now, for (b), the maximum horizontal displacement is the largest possible value for x on the circle, ie x = 20, and for (c) it's the largest possible y, ie y = 15.

 

[Xayma]

But wont the x-component stop moving when it hits the ground? As far as project component is concerned? As 20 is achieved when @=0?

 

[ND]

Originally posted by Xayma
But for b) I think that its makes distance will be when it hits the ground after 1 second. So you could work out the angle from that then stick it in the first formula.

Yep that's correct. It's a balance between horizontal velocity and not hitting the ground too early; it hitting the ground after 1 sec is optimum.

 

[CM_Tutor]

OK, if you want to look at it that way, then it's the largest x on the circle subject to y non-negative. Hence, x = sqrt(375) = 5 * sqrt(15)

 

[jimmik]

thanks! that was nicely done.