Projectile Motion - Help again

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#1
Hi guys, so im having a little trouble with the projectile motion question and if anyone could help me out it will be much appreciated!

(a) A cricket ball is thrown with a velocity of 31.4 ms-1 at an elevation angle
of 49 degrees.

Ignoring effects of air resistance, determine:

(i) The maximum height obtained.
(6 marks)

(ii) The time of flight and the range on a horizontal plane.
(5 marks)

(iii) The velocity (magnitude and direction) at a height of 18.0 m.
(5 marks)

(b) Determine the velocity of throw required at an elevation angle of 41 degrees to reach the ball catcher at -1.5 m elevation (catcher 1.5 m lower than the thrower’s arm) located at a horizontal distance of 87 m.
(9 marks)

Ok, for part (i) i used the equation H = v^2sin^2x / 2g, and got the answer to be 28.7m to 3.s.f

for part (ii) i used the equation t = 2vsinx / g, and got the answer to be 4.83s to 3.s.f
and for the horizontal range i used R = v^2sin2x / g which gave me 99.5m to 3.s.f

for part (iii) im having some difficulty with this i used the equation H = v^2sin^2x / 2g and made 'V' the subject to work out the velocity which gave me the answer to be 24.9m/s + and - and this is the speed twice - whilst going up and then coming back down again hence the signs however how do i workout the direction?

and for part B i still dont know how to do that question please can someone help me

thanks
0
6 years ago
#2
Why have you posted this question again?
You already have a thread for it where you have been geting help.
Here
http://www.thestudentroom.co.uk/show....php?t=2631571
0
#3
(Original post by Stonebridge)
Why have you posted this question again?
You already have a thread for it where you have been geting help.
Here
http://www.thestudentroom.co.uk/show....php?t=2631571
This one has more info to what i have done so far. i didn't understand the advice i got for part b of the question therefore re-posted
0
6 years ago
#4
(Original post by Dontay)
This one has more info to what i have done so far. i didn't understand the advice i got for part b of the question therefore re-posted
Ok, but it's better to keep everything in the one thread if you can. Just post again in it and it will bump back to the top.
0
#5
(Original post by Stonebridge)
Ok, but it's better to keep everything in the one thread if you can. Just post again in it and it will bump back to the top.
Uhh..help?
0
6 years ago
#6
For part iii to work out direction, take the vertical velocity component at that height* and combine it with the constant value of the horizontal component of the velocity.

* Use v²=u² + 2as

where u is the vertical component of the initial velocity.
* v is the vertical component you are looking for
s in the vertical height reached

You then have two components of velocity.
The angle is found from trig using tangent.
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