Answer:
= 308.5 N
Explanation:
acceleration of rocket for safe landing
[tex]v_{y} ^{2} = v_{0} ^{2} +2ay[/tex]
[tex]a = \frac{v_{y}^{2} - v_{0}^{2} }{2y}[/tex]
[tex]v_{0}[/tex] = initial velocity
[tex]v_{y}[/tex] =final Velocity
m = mass of rocket
[tex]\frac{0^{2} - 30^{2} }{2\times80}[/tex]
-5.625 [tex]m/s^{2}[/tex]
Upward force
F - mg = ma
F = ma+ mg
F = m(a+g)
m= mass
a = 5.625 [tex]m/s^{2}[/tex]
g = 9.8[tex]m/s^{2}[/tex]
= 20(5.625 [tex]m/s^{2}[/tex]+ 9.8[tex]m/s^{2}[/tex])
= 308.5 N
The upward force applied on the rocket is 308.5 N.
The given parameters;
mass of the rock, m = 20 kg
speed of the rock, 30 m/s
distance traveled, h = 80 m
The acceleration of the rocket is calculated by applying third kinematic equation;
[tex]v^2 = u^2 +2ah\\\\v^2 = 2ah\\\\a = \frac{v^2 }{2h} \\\\a = \frac{(30)^2}{2(80)} \\\\a= 5.625 \ m/s^2[/tex]
The upward force applied on the rocket is calculated as follows;
F = ma + mg
F = m(a + g)
F = 20(5.625 + 9.8)
F = 308.5 N.
Thus, the upward force applied on the rocket is 308.5 N.
Learn more here:https://brainly.com/question/21313437
