Question

In the first second of its flight a rocket ejects 1/60 of its mass with a velocity of 2400 m/s.the acceleration of the rocket is

Answer 1

F=v(dm/dt)
ma=2400*(1/60)m. (m is mass of rocket)
a=40m/s2

Answer 2

Hey dear,

◆ Answer-
a = 40 m/s^2

◆ Explanation-
# Given-
v = 2400 m/s
t = 1 s
m = M/60

# Solution-
Ejection force of the rocket is-
F = mv/t
F = M/60 × 2400 × 1
F = 40M

Acceleration of the rocket is calculated by-
a = F/M
a = 40M/M
a = 40 m/s^2

Therefore, acceleration of the rocket is 40 m/s^2.

Hope this helps...