A rocket moves upward, starting from rest with an acceleration of 26.3 m/s2 for 3.91 s. It runs out of fuel at the end of the 3.91 s but does not stop. How high does it rise above the ground?
To find out how high the rocket rises above the ground, we can use the kinematic equation for vertical motion:
h = h0 + v0t + 0.5at^2
Where:
h = height above the ground (what we want to find)
h0 = initial height (assumed to be 0, since the rocket starts from the ground)
v0 = initial velocity (0 m/s, since the rocket starts from rest)
t = time (3.91 s)
a = acceleration (26.3 m/s^2)
Plugging in the given values:
h = 0 + (0)(3.91) + 0.5(26.3)(3.91)^2
Simplifying further:
h = 0 + 0 + 0.5(26.3)(15.2641)
h ≈ 0 + 0 + 199.723
h ≈ 199.723
Therefore, the rocket rises approximately 199.723 meters above the ground.
To find the height the rocket rises above the ground, we can use the equations of motion for uniformly accelerated motion.
The equation we will use is:
h = (1/2) * a * t^2
where:
h = height
a = acceleration
t = time
Given:
Acceleration (a) = 26.3 m/s^2
Time (t) = 3.91 s
Substituting these values into the equation, we get:
h = (1/2) * (26.3 m/s^2) * (3.91 s)^2
Calculating this equation, we find:
h = 0.5 * 26.3 m/s^2 * (3.91 s)^2
h = 0.5 * 26.3 m/s^2 * 15.2881 s^2
h = 199.658 m
Therefore, the rocket rises approximately 199.658 meters above the ground.