A body is projected with a speed of 40 m/s vertically up from the grou

1.	A body is projected with a speed of 40 m/s vertically up from the ground. What is the maximum height reached by the body? What is the entire time of motion? What is the velocity at 5 seconds after the projection? Take g=10m/s² (Ans: 80m; 8s; 10m/s down)
Answer» Initial VELOCITY of the body(u) = 40 m/s. Final Velocity(v) = 0 Acceleration due to GRAVITY = 10 m/s². ∴ Acceleration of the body = -10 m/s². [∵ The body is thrown against the gravity]. Using the Galilio's Third equation of the motion, v² - u² = 2aS ⇒ (0)² - (40)² = 2(-10)S 1600 = 20 S ∴ S = 80 m. Hence, the maximum height reached by the body is 80 m. Now, for the time, Using the Second equation of motion, S = UT + 1/2 at² ⇒ 80 = 40t - 5t² ∴ 5t² - 40t + 80 = 0 ∴ t² - 8t + 16 = 0 ⇒ t² - 4T - 4t + 16 = 0 ∴ t(t - 4) - 4(t - 4) = 0 ⇒ (t - 4)(t - 4) = 0 ∴ t = 4 seconds. Now, If the body is projected upwards, it must be coming downwards. ∴ Total time = Time of ascent + time of descent. = 4 + 4 [Time of ascent = time of descent]. = 8 seconds. Time of ascent is 4 seconds. Thus, body must be falling down at the 5 th second. We need to find the Final velocity of the body during the fall at t = 5 seconds. In 5th second, the velocity decreases by an amount 5 × 10 = 50 m/s. Hence after 5 second it’s velocity is 40 - 50 = - 10 m/s Negative sign shows opposite direction,i.e. downwards. Hope it helps.

A body is projected with a speed of 40 m/s vertically up from the ground. What is the maximum height reached by the body? What is the entire time of motion? What is the velocity at 5 seconds after the projection? Take g=10m/s² (Ans: 80m; 8s; 10m/s down)

Answer»

Initial VELOCITY of the body(u) = 40 m/s.

Final Velocity(v) = 0

Acceleration due to GRAVITY = 10 m/s².

∴ Acceleration of the body = -10 m/s².

[∵ The body is thrown against the gravity].

Using the Galilio's Third equation of the motion,

v² - u² = 2aS

⇒ (0)² - (40)² = 2(-10)S

1600 = 20 S

∴ S = 80 m.

Hence, the maximum height reached by the body is 80 m.

Now, for the time, Using the Second equation of motion,

S = UT + 1/2 at²

⇒ 80 = 40t - 5t²

∴ 5t² - 40t + 80 = 0

∴ t² - 8t + 16 = 0

⇒ t² - 4T - 4t + 16 = 0

∴ t(t - 4) - 4(t - 4) = 0

⇒ (t - 4)(t - 4) = 0

∴ t = 4 seconds.

Now, If the body is projected upwards, it must be coming downwards.

∴ Total time = Time of ascent + time of descent.

= 4 + 4 [Time of ascent = time of descent].

= 8 seconds.

Time of ascent is 4 seconds.

Thus, body must be falling down at the 5 th second.

We need to find the Final velocity of the body during the fall at t = 5 seconds.

In 5th second, the velocity decreases by an amount 5 × 10 = 50 m/s.

Hence after 5 second it’s velocity is 40 - 50 = - 10 m/s

Negative sign shows opposite direction,i.e. downwards.

Hope it helps.