Question

a stone is allowed to fall from the top of tower 100m high and at the same time another stone is thrown upward with speed 25m/s calculate when and where the two stone meet​

Answer 1

Answer:

(100-S) = distance travelled by the projected stone. Therefore, Two stones will meet after 4 s. = 80 m. Thus , both the stone will meet at a distance of 80 m from the top of tower.

Answer 2

Explanation:

Given :-

▪ Energy of a ball falling from a height of h = 10 m is reduced by 40% when it strikes the surface.

To Find :-

▪ How high will it rebound.

Solution :-

Let the mass of the ball be m. So, when it was at a height of h = 10 m, It had no kinetic energy but only potential energy.

⇒ Potential Energy = mgh

⇒ P.E = m × 10 × 10 [ g = 10 m/s² ]

⇒ P.E = 100m J

Now, When it stroke the ground it had no potential energy but only kinetic energy which is equal to 100m J (According to the conservation of energy)

Now, It is given that the energy of the ball was reduced by 40% when it stroke the ground, So after striking the ground it had ,

⇒ k.e = K.E - 40% of K.E

⇒ k.e = mgh - 40mgh / 100 [ K.E = mgh ]

⇒ k.e = 100m - 40×100m/100

⇒ k.e = 100m - 40m

⇒ k.e = 60m J

Now, According to the conservation of energy, The kinetic energy will be converted to potential energy, So

⇒ Potential energy = Kinetic energy

⇒ mgh = 60m

⇒ 10m × h = 60m [ g = 10 m/s² ]

⇒ 10h = 60

⇒ h = 6

Hence, The ball will rebound to a height of 6 m.