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Answer :
Answer:
a) The initial acceleration is 7.84 m/s²
b) The impact speed is 117.6 m/s
c) The height is 705.6 m
Explanation:
a) The speed from A to B is:
v = u + at
Where
u = initial speed = 0
t = 10 s
Replacing:
v = 10t (eq. 1)
The vertical distance between A to B is:
[tex]h=\frac{1}{2} at^{2} +ut=\frac{1}{2} a*(10)^{2}+0=50a[/tex] (eq. 2)
From B to C, the time it take is equal to 20 s, then:
[tex]h=vt+\frac{1}{2} at^{2}[/tex]
Replacing eq. 1 and 2:
[tex]-50a=(10a*20)-\frac{1}{2} *g*20^{2} \\-250a=-\frac{1}{2} *9.8*20^{2} \\a=7.84m/s^{2}[/tex]
b) The impact speed is equal:
[tex]v_{i} ^{2} =v^{2} +2gs[/tex]
Where
s = h = -50a
[tex]v^{2} _{i} =(10a)^{2} +2*(-9.8)*(-50a)\\v=\sqrt{100a^{2}+980a } \\v=\sqrt{(100*7.84^{2})+(980*7.84) } =117.6m/s[/tex]
c) The height is:
[tex]v_{i} ^{2} =v^{2} +2gs\\0=(10a)^{2} -2gs\\(10a)^{2} =2gs\\s=\frac{(10a)^{2} }{2g} \\s=\frac{(10*7.84)^{2} }{2*9.8} =313.6m[/tex]
htotal = 313.6 + 50a = 313.6 + (50*7.84) = 705.6 m
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Answer:
Explanation:
Let a be the acceleration of launch. In 10 seconds , Distance gone up in 10 seconds
s = 1/2at²
= .5 x a x 100
=50a
Velocity after 10 s
u = at = 10a
Now 50a distance in downward direction is travelled in 20 s with initial velocity 10a in upward direction.
s = ut + 1/2 gt²
50a = -10ax20 + .5 x 10x400
250a = 2000
a = 8 m s⁻² .
Velocity after 10 s
= at = 80 m/s
further height reached with this speed under free fall
h = v² / 2g
= 80 x 80 / 2 x 10
= 320 m
height achieved under acceleration
= 50a
= 50 x 8 = 400m
Total height
= 320 + 400= 720 m
velocity after falling from 720 m
v² = 2gh = 2 x10 x 720
v = 120 m/s
