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astarmathsandphysics:
thrid question
resolving for 2kg mass
T-2g=2a
Resolving for 5 kg mass
5g-T=5a
add these
g=5a so a=g/5
sub into T-2g=2a
T=2g+2*g/5 so T=12g/5

astarmathsandphysics:
q3

LekhB:
Thanks a lot.

From a point 120m above ground level, a stone is projected vertically upwards with a speed u m/s. if the stone rises to a height of 5m above the point of projection, calculate(using g= 10 m/s²)
(i)  the value of u,
(ii) the time the stone takes from projection until it reaches the ground level,
(iii) the speed of the stone at ground level.



A particle is projected vertically upwards from ground level. Between 2 seconds and 3 seconds after leaving the ground, it rises 45 m. Calculate (using g=10 m/s²)
(i) the speed of projection
(ii) the maximum height reached
(iii) the time interval for which the particle is above a level of 165m.

LekhB:
I hate not getting the same answers as written in my homework book but I don't find anything wrong with my calculations.

A particle is projected vertically upwards from the ground with a speed of 36 m/s. Calculate (using g= 10 m/s²)
(i) the time for which it is above a height of 63m

What I tried :

S: 63m
u: 36 m/s
v: ? m/s
a: 10 m/s²
t: ? s

S= ut + 0.5(at²)

63 = 36(t) + 0.5(10)(t²)

-> 5t² + 36t - 63 = 0

t = [(-36) + (36² - 4(5)(-63)^0.5]/10 or [(-36) - (36² - 4(5)(-63)^0.5]/10

t= 1.46 s.

The answer is t= 1.2s.

And I need t for doing the other parts which are;

(ii) the speed which it has at this height on its way down
(iii) the total time of flight.

LekhB:
Two more


From the foot of a vertical cliff 28.8 high, a stone was projected vertically upwards so as just to reach the top. Find its velocity of projection.
One second after the first stone was projected, another stone was allowed to fall from rest from the top of the cliff. The stones passed one another after a further t seconds at a height h m above the ground. Calculate the value of t and h.

Answers : 24 m/s(I got 13.78), 0.7, 26.4

I used to get all of these numbers right in physics. Don't know what happened here...

Another one(in which I got the first answer though):


A particle X is projected vertically upwards from the ground with a velocity of 80 m/s. Calculate the maximum height reached by X. -> I could do this part

The other part: A particle Y is held at a height of 300m above the ground. At the moment when X has droped 80m from its max height, Y is projected downwards with a velocity of v m/s. The particles reach the ground at the same time. Calculate the value of v.

I got v= 103.4

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