Amodel rocket is fired vertically upward with an initial velocity of 49 m/s from a tower 150 m high. (a) how long will it take for the rocket to reach its maximum height? (b) what is the maximum height? (c) how long will it take for the rocket to pass its starting point on the way down? (d) what is the velocity when it passes the starting point on the way down? (e) how long will it take for the rocket to hit the ground? rocket to hit the ground?

a)  t  =  5 sec

b)  h  =  122.5 m

c)  t  =  5 sec

d)  V  = 49 m/sec

e)  t₁  =  1.7 sec

Step-by-step explanation:  Equations for vertical projectile motion are:

  V = V(i)  -  g*t                    (1)        

Δh = V(i)*t - 1/2 g*t²        (2)    

 V² - V(i)²  =  - 2g*Δh       (3)

We have to remember that rocket was fired from 150 mt above ground level.

a) How long will it take to reach its maximum height

V = V(i)  -  g*t

At point of maximum height V = 0 then  

V(i)  =  gt    

49/9.8   = t

t  = 5 sec

b) Maximum height

 V² - V(i)²  =  - 2g*Δh     where V(i)  =  49 m/sec   and V  =  0

- V(i)²  = - 2g*Δh       Δh  =  V(i)² / 2g         g  =  9.8 mt/sec²

Δh  = (49)²/ 2*9.8

Δh  =  2401/ 19.6

Δh  =  122.5 m

c) How long will the rocket take to pass its starting point on the way down

We assume the question is how long will the rocket take to pass its starting point from the maximum height

In that case at maximum height point  V(i)  = 0

and we use equation (2)

Δh = V(i)*t - 1/2 g*t²      where  V(i)  = 0

Δh  =  1/2 *g*t²      (note: we cnage sign since now g is positive the rocket is falling down)

122.5  =  1/2 *9.8 * t²         t²  =  25      

t =  5 sec    (time from maximum height to starting point in the way down)

d ) What is speed when it passes the startin point

We use equation (1)

V = V(i)  -  g*t     knowing that  V(i)  = 0 ( note we are talking about the free falling rocket )

V  = g*t          (we have to change sign because now g is positive

V  =  9.8 * 5

V  =  49  m/sec

e) How long will take the rocket to hit the ground

Using equation (2)

Δh = V(i)*t - 1/2 g*t²         where  Δh  =  150 m

150  =  49*t + 1/2 9.8*t²

we get a second degree equation

49t²  +  4.9 t  - 150  =  0

t₁,₂  = [-4.9  ± √(4.9)² +29400  ]/98

t₁,₂  = [-4.9  ± 171.53] /98      t₁ = ( -4.9  + 171.53)/98  y t₂  = (-4.9 - 171.53)/98

we dismiss negative value t₂

then

t₁  =  1.7 sec

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