Question

A rock weighing 25-N is thrown vertically into the air from ground level. when it reaches 15.0 m above ground, it is traveling at 25.0 m/s upward. Use the work-energy to find:
a) the rock's speed just as it left the ground
b) its maximum height

Answer 1

Answer(a): V_i = 30.32 m/s (approx)

Answer(b) : h_max = 46.9 m (approx)

Let the rock's speed when it left the ground be V_i

mg = 25 N

m = 25 / 9.8 = 2.55 kg

By energy coservation,

Kinetic energy (K.E.) + Potential Energy (P.E.) = constant

0.5mv² + mgh = constant

when rock is thrown,

h = 0

mg = 25 N (given)

v= V_i

Total energy = 0.5*m* V_i² + 0 ................. (1)

When the rock is at 15 m height:

h = 15m

mg = 25 N (given)

velocity at height 15m = V₁₅ = 25 m/s (given)

Total energy = 0.5*m* V₁₅² + 25 * 15 ............(2)

At maximum height,

h = h_max

mg = 25 N (given)

velocity at height h_max = V_hmax= 0 m/s (body comes at rest)

Total energy = 0.5*m* 0² + 25 * h_max   

= 25 * h_max ............(3)

By conservation of energy,

1 = 2 = 3

From (1) and (2) :

0.5*m* V_i² + 0 = 0.5*m* 25² + 25 * 15

V_i = 30.32 m/s Answer(a)

From (1) and (3):

0.5*m* V_i² + 0 = 25 * h_max

h_max = 46.9 m Answer(b)

Please feel free to reach out for further clarifications. Thanks.