Given data:

g = 10m/s^{2}

Total time T = 6sec

T_{a} = T_{d} = 3sec

(a) Final velocity at maximum height v = 0

From first equation of motion:-

v = u - gt_{a}

u = v + gt_{a}

= 0 + 10 x 3

= 30m/s

The velocity with which the stone was thrown up is 30m/s.

(b) From second equation of motion

\( s = ut_a - \frac{1}{2}g(t_a)^2\)

\( = 30 \times 3 - (1/2) \times 10 \times (3)^2\)

\( = 90-45 = 45m\)

The maximum height stone reaches is 45 m.

(c) In 3sec, it reaches the maximum height.

Distance travelled in another 1sec = s’

\( s = ut_a - \frac{1}{2}g(t_a)^2\)

\( s = 0 + 10 \times 1 \times 1 \)

s = 5m.

The distance travelled in another 1sec = 5m.

Therefore in 4sec, the position of point p (45 – 5)

= 40m from the ground.

Answered by Shivani Kumari | 1 year agoAn object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?

Illustrate the law of conservation of energy by discussing the energy changes

which occur when we draw a pendulum bob to one side and allow it to oscillate. Why

does the bob eventually come to rest? What happens to its energy eventually? Is it a

violation of the law of conservation of energy?

The volume of a 500 g sealed packet is 350 cm^{3} . Will the packet float or sink in water if the density of water is 1 g cm^{-3} ? What will be the mass of the water displaced by this packet?

The volume of 50 g of a substance is 20 cm^{3} . If the density of water is 1 g cm^{-3} , will the substance float or sink?

Why a block of plastic does released under water come up to the surface of water?