The important thing to note here is the direction of motion of the test rocket. Since it mentions that the rocket travels vertically upwards, then this motion can be applied to rectilinear equations that are derived from Newton's Laws of Motions.These useful equations are:
y = v₁t + 1/2 at²
a = (v₂-v₁)/t
where
y is the vertical distance travelled
v₁ is the initial velocity
v₂ is the final velocity
t is the time
a is the acceleration
When a test rocket is launched, there is an initial velocity in order to launch it to the sky. However, it would gradually reach terminal velocity in the solar system. At this point, the final velocity is equal to 0. So, v₂ = 0. Let's solve the second equation first.
a = (v₂-v₁)/t
a = (0-30)/t
a = -30/t
Let's substitute a to the first equation:
y = v₁t + 1/2 at²
49 = 30t + 1/2 (-30/t)t²
49 = 30t -15t
49 = 15 t
t = 49/15
t = 3.27 seconds
Answer:
15 because 5×5×5 is the same thing as 5×3 which equals to 15
Answer:bill 5 m/s. Jack:10 m/s
Explanation:
Cuz I took it
Answer:
(a) 62.5 m
(b) 7.14 s
Explanation:
initial speed, u = 35 m/s
g = 9.8 m/s^2
(a) Let the rocket raises upto height h and at maximum height the speed is zero.
Use third equation of motion
h = 62.5 m
Thus, the rocket goes upto a height of 62.5 m.
(b) Let the rocket takes time t to reach to maximum height.
By use of first equation of motion
v = u + at
0 = 35 - 9.8 t
t = 3.57 s
The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.
