M2- Projectiles help

17) A stone is projected from a point O on a cliff with a speed of 20m/s at an angle of elevation of 30 degrees. t seconds later the angle of depression of the stone from O is 45 degrees. Find the value of t.

Reply 1

Sub Zero

14) split it into horizontal and vertical components and SUVAT the horizontal and vertical.

17) draw a diagram but similar to 14 in terms of SUVAT that. just remember that displacement (s) can be negative.

Reply 2

Newcastle456

Zohair09

I know all that, but I don't know how to do it.

14) Horizontally:
s=4
a=0
t=2

Vertically:
s=4
a=-9.8
t=2

17)
Horizontally:
u=20cos30
a=0

Vertically
a=-9.8
u=20sin30

Reply 3

+ polarity -

For question 14 you can see that the angle of projection is 45 degrees, so you can split up the velocity into horizontal and vertical components, I think.

Oh! and you can use the time and the displacement to find the components of the velocity! :excited:

Reply 4

samir12

14) You can find the initial horizontal speed by using t=2 and horizontal distance = 4m. The same for the initial verticle speed only that your using the formula s = ut + 1/2at^2, bung in the verticle distance (4m) and time, and rearrange to find u (initial verticle speed).

Than just do tantheta = verticle speed / horizontal speed, and find theta which will give you your direction of motion.

The range is quite simple, just find the time it takes for the ball to reach the ground by making s = ut + 1/2at^2 equal to 0, at that time the ball has reached its range (horizontal distance) so jus sub in that t value into s = ut.

Reply 5

Newcastle456

+ polarity -

It doesn't match the answer in the back of the book of 4.82m at 80degrees.

Reply 6

Newcastle456

samir12

Horizontally:
4=2u
u=2

Vertically:
4=2u-4.9*4
4=2u-19.6
2u=23.6
u=11.8

tan feta = 11.8/2

feta=80.3 degrees.

I don't understand the next bit though.

Reply 7

+ polarity -

Deep456

It doesn't match the answer in the back of the book of 4.82m at 80degrees.

Oh sorry! I totally forgot to use the equations of motion. Just ignore my posts. :colondollar:

Right and 17?

Lou Reed

Split V into horizontal and vertical components and equate.

Vh = 10root3
Vv = |10 - 9.8t|

10root3 = -10 + 9.8t
t = 0.75 seconds

I think that's it...what's the answer in the book?

note: i used modulus on Vv so we get a positive t

Nope it is not the answer. 5.59 s is the answer.

Reply 10

Newcastle456

This question is driving me mad.

Reply 11

gcseeeman

Lou Reed had the right idea, but was trying to work out when the speed was acting at 45 degrees (although I think they made a mistake with the modulus etc because I believe that would be after ~2.79 seconds).

Angle of depression = angle between horizontal and the line from O to the particle (below horizontal). Drawing a diagram, you should see that when the angle of depression is 45 degrees, (vertical distance) / (horizontal distance) = tan (45). In this case this is relatively easy, because - as tan (45) = 1 - they are the same.

So find the horizontal distance at time T, the vertical distance (downwards) at time T, and equate them then solve for T.

Spoiler

Note I get 5.58, so I'm guessing your book has a slight rounding error? (unless I've made a mistake somewhere)