(Original post by boromir9111)
I will just put all my mechanics question into this thread rather than creating mini ones, saves me a lot of time.
For now i just have i three questions i need help on.
"The picture shows Romeo(use your imagination lol) trying to attract Juliet's attention without her nurse, who is in a downstairs room, noticing. He stands 10m from the house and lobs a small pebble at her bedroom window. Romeo throws the pebble from a height of 1m with a speed of 11.5 ms^-1 at an angle of 60 degrees to the horizontal.
(i) How long does the pebble take to the house? t = 1.74 secs(correct)
(ii)Does the pebble hit Juliet's window, the wall of the house or the downstairs room window? s = 3.50m, Juliet's window(correct)
(iii) What is the speed of the pebble when it hits the house?
This is where i get stuck, the answer is v = 9.12ms^-1 and tried splitting it up into components Vy and Vx, Vy = 11.5sin60 - 9.8*1.74(which is wrong) and Vx doesn't work, so what do i do here???
Use g = 10 ms^-2......A firework is buried so that its top is at ground level and it projects sparks all at a spped of 8ms^-1. Air resistance may be neglected.
(i) Calculate the height reached by a spark projected vertically and explain why no spark can reach a height greater than this. v^ = u^2 + 2as....s = 3.27m
(ii) For a spark projected at 30 degrees to the horizontal over horizontal ground, show that its height in metres t seconds after projection is 4t-5t^2 and hence calculate the distance it lands from the firework.
s = ut+1/2at^2, 8cos60 = 4 and so on......when it lands y = 0, so solve for t, t = 4/5......s = ut(horizontal) so, 8sin60*4/5 = 5.5m(correct)
(iii) For what angle of projection will a spark reach a maximum height of 2 m??
......i am not sure what to do here tbh, any hint would be useful
A small stone is projected horizontally from a height of 19.6m over horizontal ground at a speed of 15ms^-1
(i) Calculate the time which elapses between projection and the stone hitting the ground and show that the horizontal distance travelled is 30m.....So = 19.6m, Sy = ut+1/2at^2 = -4.9t^2
-19.6/-4.9 = t^2 = t = 2, Sx = 15*2 = 30m
In a new situation two small stones are projected horizontally towards each other at the same instant over horizontal ground. Initially the stones are 19.6m above the ground and are 50m apart. The initial speeds are 15ms^-1 and v ms^-1.
(ii) Explain briefly why the stones will be at the same height as one another above the ground after the elapse of any time interval and hence much collide if they do not hit the ground first. Y1 = Y2 = 19.6-4.9t^2 which is the same for all time intervals (correct)
(iii) Find the value of v if the stones collide 4.9m above the ground.
i got no idea what i am supposed to be doing here???
Thanks in advance
u is initial velocity, v is final v_x represent x component etc.
u_x= u cos60= v_x
10/ ucos60 = T
v_y= v_y - gT
(v_y^2 + v_x^2)^0.5= final speed
that looks like what you have, and if its wrong, I think possible the answer itself is wrong.
for q2 you know the absolute speed of firing out. from the first part, you just do the same:
max height h, projection speed u, final speed=0, projection angle Q measured from horizontal
0= u^2 sin^2(Q) -2gh
solve for Q
for q3 you can form equations for its vertical and horizontal position. becuase you know:
vertical displacement of Stone 1= vertical dis of stone 2
horizontal displacement of stone 1= hor. dis. stone 2
and the time at which this occurs has to be the same.
Therefore you could form 4 equations for each stones displacement as a fucntion of time. Then set the hor. equal and vert. equal, solving for t