How long till it hits the ground A model rocket is fired int

How long till it hits the ground?

A model rocket is fired into the air. Over the first four seconds, the rocket\'s engines give it a velocity of 51 m/s, where t is the time since liftoff. Over the next five seconds, the rocket continues to move vertically but is now pulled downwards by gravity, so has velocity 40 - 5t. Finally, a parachute unfurls and the rocket falls at velocity -10 + 5e-(t-9) for all t 9, until the rocket lands.

Solution

For first 4 s, acceleration = d(5t)/dt = 5 m/s^2

Distance travlled in first 4 s is, 1/2*5*4^2 = 40 m

velocity after first 4 s is, 5*4 = 20 m/s

For next 5 s, acceleration = d(40-5t)/dt = -5 m/s^2

Velocity will be zero when 0 = 40-5t

This gives t = 8 s

Distance travelled from 4 s to 8 s is 20*(8-4) - 1/2*5*(8-4)^2 = 40 m

Distance travelled from 8 s to 9 s is -1/2*5*(9-8)^2 = -2.5 m

Velocity at 9 s is 40-5*9 = -5 m/s

Distance from the ground after 9 s is 40 + 40 - 2.5 = 77.5 m

Velocity after 9 s is, v = dx/dt = -10 + 5e^(-(t-9))

So, dx = [-10 + 5e^(-(t-9))]dt

Integrating both the sides, x = -10t - 5e^(-(t-9))

For the period from 9 s to the time it touches the ground, -77.5 = -10t - 5e^(-(t-9)) - [-10*9 - 5e^(-(9-9))]

Or, -77.5 = -10t - 5e^(-(t-9)) + 95

t = 17.25 s