Question

8.22) A stone is thrown vertically
up ward with an initial velocIty of
40ms_1 Taking 9 = 10ms_1? Find the maximum height reached by the stone. what is the net displacement and the
label distance covered by the stone?
​

Answer 1

Answer:

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v 2

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v 2 =u

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v 2 =u 2

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v 2 =u 2 −2as gives v

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v 2 =u 2 −2as gives v 2

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v 2 =u 2 −2as gives v 2 =u

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v 2 =u 2 −2as gives v 2 =u 2

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v 2 =u 2 −2as gives v 2 =u 2 −2aH, we have,

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v 2 =u 2 −2as gives v 2 =u 2 −2aH, we have,H = 2gu 2−v 2=2(10m/s 2)40m/s

Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.Acceleration due to gravity, g = 10m/s 2 (downward motion).Maximum height, s = H.As the body is thrown upward a = -g the relation v 2 =u 2 −2as gives v 2 =u 2 −2aH, we have,H = 2gu 2−v 2=2(10m/s 2)40m/s 2 −0 2= 20/1600=80m.

1600=80m.If a stone is thrown vertically upward, it returns to its initial position after achieving maximum height.

1600=80m.If a stone is thrown vertically upward, it returns to its initial position after achieving maximum height.so, the net displacement = Difference of positions between initial and final positions = 0.

1600=80m.If a stone is thrown vertically upward, it returns to its initial position after achieving maximum height.so, the net displacement = Difference of positions between initial and final positions = 0.Total distance covered = 80 m + 80 m = 160 m.

1600=80m.If a stone is thrown vertically upward, it returns to its initial position after achieving maximum height.so, the net displacement = Difference of positions between initial and final positions = 0.Total distance covered = 80 m + 80 m = 160 m.Hence, the displacement is 0 and the total distance covered is 160 m.

Answer 2

Correct question :

A stone is thrown vertically  upward with an initial velocIty of  40ms⁻¹ Taking g = 10ms⁻² . Find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Given :

• IntiaI veIocity = 40 ms⁻¹
• AcceIeration due to gravity = 10 ms⁻²

To find :

• Maximum height reached by stone
• Net dispIacement
• TotaI distance covered

SoIution :

As we have u = 40 ms⁻¹ and g = 10 ms⁻²

We know the formuIa for maximum height reached :

⇔ Maximum height (h) = u²/2g

⇒ h = (40)²/(2 * 10)

⇒ h = 1600/20

⇒ h = 80 m

∴ Maximum height reached by stone = 80 m

Now net dispIacement :

Stone was thrown upwards and it came back to ground after reaching maximum height so it came back to it's initiaI position. We know that if a body covers a particuIar distance and then comes back to it's initiaI position so it's net dispIacement = 0.

⇒ DispIacement = 80 - 80

⇒ DispIacement = 0

∴ Net dispIacement of stone = 0

Now totaI distance :

As body comes back to it's initiaI position so it covers same distance twice :

⇒ TotaI distance = 80 + 80

⇒ TotaI distance = 160 m

∴ TotaI distance covered = 160 m