An experimental rocket designed to land upright falls freely from a height of 2.88 102 m, starting at rest. At a height of 95.9 m, the rocket's engines start and provide constant upward acceleration until the rocket lands. What acceleration is required if the speed on touchdown is to be zero? (Neglect air resistance.)

Vf^2=Vi^2+2Anet*d where

Anet=-9.8+a

Now vi, figure from change of PE
1/2 m vi^2=mg*(-288)-ma(-95.9)
solve for vi.

Then,
vf^2=0=Vi^2+2*-9.8*-95.9+2a*-95.9
solve for a

Note Vi is negative, but when squared, postive. a should end up to be positive. Notice and think out why I put a - on the 288, on the 9.8, and on the 95.6.

What acceleration is required if the speed on touchdown is to be zero?

To solve this problem, we can use kinematic equations of motion. Let's break down the information given:

Initial height (s0) = 2.88 * 10^2 m
Final height (s) = 95.9 m
Initial velocity (v0) = 0 (starting at rest)
Final velocity (v) = 0 (speed on touchdown)

Since the rocket is falling freely from a height, its initial acceleration (a0) is equal to the acceleration due to gravity (g), which is approximately 9.8 m/s^2.

We need to find the acceleration (a) required when the rocket's engines start to decelerate it before landing. At this point, the acceleration is constant until the rocket lands.

To find the acceleration, we can use the following equation of motion:

v^2 = v0^2 + 2a(s - s0)

Plugging in the given values, we get:

0^2 = 0^2 + 2a(95.9 - 2.88 * 10^2)

Simplifying the equation further:

0 = 2a(95.9 - 2.88 * 10^2)

0 = 191.8a - 2a * 2.88 * 10^2

Rearranging the equation to solve for acceleration:

2a * 2.88 * 10^2 = 191.8a

Simplifying further:

577.6a = 191.8a

Dividing both sides of the equation by 191.8:

577.6 = a

Therefore, the required acceleration to achieve a speed of zero on touchdown is approximately 577.6 m/s^2.