Question

A rocket engine applies a thrust for e fo 450N for 15s. The 2800Kg rocket is travelling 690 m/s before the engine thrust. What is the final velocity of the rocket?

Answer 1

692.4m/s

Step-by-step explanation:

According to Newton's second law;

F = ma

F = m(v-u)/t

m is the mass

v is the final velocity

u is the initial velocity

t is the time

F is the forcr

Given

F = 450N

m = 2800kg

u =690m/s

t =15s

v = ?

Substitute into the formula

450 = 2800(v-690)/15

450×15 = 2800(v-690)

6750 = 2800(v-690)

6750/2800 =v-690

2.4 = v-690

v = 690+2.4

v = 692.4m/s

Hence the final velocity is 692.4m/s