Answer:
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
