Question

A drone has been designed that can do monitoring and surveillance at considerable heights due to its light weight of 0.800 kg. For this purpose, they are doing a test to determine its maximum height and they make it start in a vertical direction, using its thrusters it manages to achieve a thrust of 35.0 N during the first 6.00 s. until the battery runs out. What was the maximum height that the drone reached?

Answer 1

2660 m

Step-by-step explanation:

Sum of the forces in the first 6.00 s:

∑F = ma

F − mg = ma

35.0 N − (0.800 kg) (10 m/s²) = (0.800 kg) a

a = 33.75 m/s²

The height it reaches during this time is:

Δy = v₀ t + ½ at²

Δy = (0 m/s) (6.00 s) + ½ (33.75 m/s²) (6.00 s)²

Δy = 607.5 m

The velocity it reaches is:

v = at + v₀

v = (33.75 m/s²) (6.00 s) + 0 m/s

v = 202.5 m/s

After the battery runs out, the drone is in free fall. At the highest point, the velocity is 0. The height at this point is:

v² = v₀² + 2aΔy

(0 m/s)² = (202.5 m/s)² + 2 (-10 m/s²) (h − 607.5)

h ≈ 2660 m