Question

A model rocket accelerates upward from the ground with a constant acceleration, reaching a height of 1250 m in 10 s. a) What is the speed (in m/s) at a height of 1250 m? 2.50×102 b) What is the acceleration (in m/s 2)?

Answer 1

Let Initial velocity be u m/s

Height upto which rocket reached,h = 1250m

time taken by the rocket,t= 10s

Accelaration due to gravity, g= 9.8 m/s²

Let accelaration of the rocket be a m/s²

Net accelaration experienced by the rocket= (a-g) m/s²

(b) Accelaration of the rocket= a m/s²

Now, as per equation of motion,

h= ut+ 1/2 (a-g)* t²

1250= 1/2* (a-9.8)* 100

2,50,000= (a-9.8)

a= (250,000+9.8) m/s²

a= 250,009.8 m/s²

a= 2.5* 10⁵ m/s²

a= 2.5* 10² km/s²

(a) Final velocity of Rocket, v m/s

v= u+ at

v= 2.5* 10²* 10 km/s²

a= 2.5* 10³ km/s²