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)
Answers
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.
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.
Explanation:
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