A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 26.0 m/s, and the air drag on it is 0.262 N throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground

a) 19.4 m/s

b) 19 m/s

Explanation:

a) In the given question,

the potential energy at the initial point = Ui = 0

the potential energy at the final point = Uf = mgh

the kinetic energy at the initial point = Ki = 1/2 mv₀².

the kinetic energy at the final point = Kf = 0

work done by air= Ea= fh =  0.262 N

Now, using the law of conservation of energy

initial energy= final energy

Ki +Ui = Kf + Uf +Ea

1/2 mv₀² + 0 = 0 + mgh + fh

1/2 mv₀² = mgh + fh

h = v₀²/ 2g (1 +f/w)

calculate m

m= w/g = 5.29 /9.8

= 0.54 kg

h = 20 ²/ (2 x9.80) x (1 0.265/5.29)

h = 19.4 m.

b) 1/2 mv² + 2fh = 1/2 mv₀²

Vg = 19 m/s

c

explanation: