Question

❒ Moderators
❒ Brainly Stars
❒Other best Users


${\large{\pink{➢{\blue{\underline{\red{\textbf{Question :}}}}}}}}$
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g= 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

${\orange{\underline{\bf{Note :-}}}}$
⊙ Quality answers needed.
⊙ Spams will be reported .​

Answer 1

Given

• Initial velocity = 40 m/s
• g = 10 m/s²

To Find

• Maximum height
• Distance & Displacement

Solution

☯ H = u²sin²θ/2g

• Here we shall use the value of θ as 90° so sin 90 = 1

━━━━━━━━━━━━━━━━━━━━━━━━━

✭ According to the Question :

➞ H = u²sin²θ/2g

➞ H = (40²×1)/2(10)

➞ H = 1600/20

➞ H_max = 80 m

━━━━━━━━━━━━━━━━━━━━

• Distance = 2 × 80 = 160 m (Because 80 m travelling up and 80 m down)
• Displacement = 0 m (Because it comes back to its initial position)

Answer 2

Question:

• A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Answer:

• Maximum height = 80 m
• Total distance = 160 m
• Net displacement = 0 m

Explanation:

Given:

• Final velocity (v) = 0 m/s
• Initial velocity (u) = 40 m/s
• Acceleration (a) = g = 10 m/s²

To Find:

• Distance covered by stone while going upward i.e, height of stone?
• Net displacement?
• Total distance covered?

Solution:

Using third equation of motion. As we know that,

• v² = u² + 2as

Here, acceleration will be -10 m/s² as the acceleration is opposite to the motion of ovject.

Putting all values we get,

➡ 0² = 40² + 2(-10)(s)

➡ 0 × 0 = 40 × 40 + 2 × -10 × s

➡ 0 = 1600 + (-20s)

➡ 0 = 1600 - 20s

➡ 0 - 1600 = -20s

➡ -1600 = -20s

➡ s = -1600/-20

➡ s = 1600/20

➡ s = 160/2

➡ s = 80 m

Hence, maximum height reached by stone is 80 m.

Now, finding total distance covered by stone i.e, distance covered by stone while going upward and coming down, it will be given by sum of distance while going up and distance while coming down and both distances are same so,

➡ Total distance = 80 + 80

➡ Total distance = 160 m

Now, here net displacement will be zero as the object came back to its initial position.

▬▬▬▬▬▬▬▬▬▬▬▬