A rocket is launched straight up with a velocity of 4.26 m/s. What would be the velocity when it lands?

Answer 1

Hello Pamela -

Thanks for asking this question! It turns out that the rocket will have exactly the same speed when it lands as when it takes off, just going down instead of up - this is because projectile motion (i.e. the motion of an object launched into the air where the only force on it is gravity) is symmetrical. One way to see this is to use the 1-dimensional kinematics equation that models the motion of an object undergoing acceleration, given first in words and then in symbols:

change in y position = 1/2 * acceleration in y direction * (time)² + initial velocity in y direction * time

Δy = 1/2 * a * t² + v₀ * t

We know that this rocket is accelerating at -9.8 m/s² due to gravity, and we are given its initial velocity, so the equation looks like this with known values plugged in:

Δy = 1/2 * -9.8 * t² + 4.26 * t

As we can see, there are two unknowns in this equation: the rocket's change in position, and the time it will be in the air for. However, if we think about the two moments that we care about most in this problem, we can figure out the value for Δy - we are given information about the rocket at its moment of takeoff, and we want to know the rocket's velocity at the moment that it hits the ground, so Δy between these two moments will be zero since the rocket is in the same position: just leaving the ground at the beginning, and just touching down at the end. That means we can set Δy equal to 0 in the equation above and solve for the value of t:

0 = 1/2 * -9.8 m/s² * t² + 4.26 m/s * t

0 = -4.9t² + 4.26t

t = 0 sec., t = 0.869 sec.

The quadratic equation above can be solved by graphing or applying the quadratic formula (let me know if this is confusing to you), and we find that Δy = 0 for t values of 0 and 0.869, meaning that the rocket will be at its initial position at time 0 (before it launches) and at time 0.869 seconds from launch (when it is touching down).

Now that we know the rocket's flight time of 0.869 seconds, to find the velocity at touchdown, we use the kinematics equation for the change in velocity of an object undergoing acceleration over time:

final velocity = initial velocity + (acceleration * time)

v_f = v₀ + (a*t)

v_f = 4.26 m/s + (-9.8 m/s² * 0.869 s)

v_f = -4.26 m/s

The final velocity of the rocket will be -4.26 m/s - the same speed it began with, but now going in the opposite direction. Let me know if any questions and I'll be happy to explain more! Hope this helps.