A plane fires a rocket of mass 70kg horizontally with velocity 150m/s from hight 800m from ground, it hits the ground and makes a hole of depth 4m then it stops , find the potential energy of the rocket at instant of releasing , the velocity when it reaches the ground and the average resistance of earth to the motion of the rocket.
Answers
Explanation:
i... potential energy at the instant of releasing = MGH=70 X 9.8 x 800=548800 joules
ii... Velocity when it reaches the ground = u square + 2 GH = 0 square + 2*9.8 * 800 =15680 metre per second
iii... Resistance offered by earth will be given by third equation of motion that is V^2=U^2 + 2 GH
We need to find G , so G= -U^2/2H
= -15680^2/2*4= -614 65600 metre per second square
Or the resistance offered= +614 65600 metre per second square
Answer:1.4×10^5
Explanation:
i) instant potential energy = mgh (where g=10m/s^2)
= (70)(10)(800)=5.6×10^5
ii) (P.E)i = (K.E)f
5.6×10^5 = 1/2 m(vf^2) = 35(vf^2)
Vf(when it reached the ground) =127(approximately)
iii) from reachingthe ground till it stops.
Vf=0
Vi=127m/s
d=4m
m=70kg
F=ma
Vf^2=Vi^2+2ad (since Vf=0)
a= - Vi^2 / 2d
F (resistance) = (m) (-Vi^2/2d)= -1.4×10^5