The potential energy of the spring is 6.75 J
The elastic potential energy stored in the spring is given by the equation:
where;
k is the spring constant
x is the compression/stretching of the string
In this problem, we have the spring as follows:
k = 150 N/m is the spring constant
x = 0.3 m is the compression
Substituting in the equation, we get
Therefore. the elastic potential energy stored in the spring is 6.75J .
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Answer:
time rising = 34 / 9.8 = 3.47 sec
total time in air = 2 * 3.47 sec = 6.94 sec
(time rising must equal time falling)
R = 17 m/s * 6.94 s = 118 m
Can also use range formula
R = v^2 sin (2 theta) / g
tan theta = 34 / 17 = 2
theta = 63.4 deg
2 theta = 126.9 deg
sin 126.9 = .8
v^2 = 17^2 + 34^2 = 1445 m^2/s^2
R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above
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Answer:
= 308.5 N
Explanation:
acceleration of rocket for safe landing
= initial velocity
=final Velocity
m = mass of rocket
-5.625 
Upward force
F - mg = ma
F = ma+ mg
F = m(a+g)
m= mass
a = 5.625
g = 9.8
= 20(5.625 + 9.8)
= 308.5 N
