Question

Q1. a rocket launched accelerates at 3.5 m/s^2 in 5.90 secs and 2.98 m/s^2 in the next 5.98 secs and then experiences a free fall. what time will the rocket be in air? assume that the rocket is launched from the ground.

Answer 1

CALCULATIONS

s₁ = 0.5 × 3.5 × 5.9² = 60.9 m 

V₁=at = 3.5 × 5.9 = 20.65 m/s. :

Velocity after 3.5 s. 

S₂ = Ut + 0.5 at²
 
S₂ = 20.65 × 5.98 + 0.5² × 2.98 × 5.98² = 176.8 m. 

V₂ = at = 2.98 × 5.98 = 17.82 m/s. 

Velocity at 176.8 m. 

S₃ = (V² - u²) / 2g  ( maximum height )

S₃ = (0-17.82²) /-19.6 = 16.2 m.

This is free fall distance upwards.
 
t  ( Time to reach maximum height) = (v - u)/g

g = 9.8
 
 = (0 - 17.82) / (-9.8) = 1.82 s. 

Time to reach maximum  Height from the point it is released from the ground
 
Tr (Sum all the times up to maximum height ) = 5.90+5.98+1.82 = 13.7 s.
 
Total height  = S₁ + S₂ + S₃
 
H = 60.9 + 176.8 + 16.2 = 254 m 

Fall time equals :

H = ut + 0.5 gt² = 254 m.
 
0 + 4.9 t² = 254 

t² = 51.8 

Tf = 7.2 s. = Fall time. 

Time in air = Tr (Time to reach maximum height) + Tf (Fall tme) = 13.7 + 7.2 = 20.9 s

This equals to time in air.