Question

a ball is thrown vertically upward from the ground at time t=0s it passes through the top of a tower t=3s and 2s layer it reaches its maximum height the height of the tower is​

Answer 1

Answer: 105 m

Solution:

A ball thrown vertically upward from the ground given it's free fall with a velocity of u reaches its maximum height in t seconds.

Then, the maximum height achieved is given by the Newton's third equation of motion.

• 2gh = v² - u² ( here, a = g )

In the question, it is given that the ball reaches the top of the tower at t = 3s, and at t = 5s, it reaches its maximum height.

(It reaches to the top at t = 5 because, It is given in the question that the ball reaches its maximum height 2 seconds later when it is at the top of the tower.)

Now, we have

• t = 5s, v = 0s , g = -10 m/s²

( v = 0, because at its maximum height, the velocity of the would be zero since the ball won't be moving any further, and from this point, the ball will start falling. )

( Also, g = -10 m/s², you may notice the negative sign, It's because the direction of the acceleration is opposite to the motion of the ball. )

So, let's find the initial velocity, v using the Newton's first equation of motion

> v = u + gt

> 0 = u + (-10)×5

> 0 = u - 50

> u = 50 m/s

Now that we have initial velocity, let's find the velocity at t = 3s because at this Interval of time, the ball is at the top of tower.

> v' = u + at

> v' = 50 + (-10) × 3

> v' = 50 - 30

> v' = 20 m/s

Substituting the required values in the third equation of motion, the height of the tower can be found in the following way,

> 2gh = v'² - u²

> 2×-10×h = (20)² - (50)²

> -20h = 400 - 2500

> -20h = -2100

> 20h = 2100

> h = 210/2

> h = 105 m

Hence, The height of the tower is 105m.

Answer 2

Question:

• A ball is thrown vertically upward from the ground at time (t) = 0s it passes through the top of a tower at time (t) = 3s and at time (t) = 2s it reaches it's maximum height. The height of the tower is?

Answer:

• The height of the tower is 105m.

Explanation:

Given that:

• A ball is thrown vertically upward from the ground at time (t) = 0s
• It passes through the top of a tower at time (t) = 3s and,
• At time (t) = 2s it reaches its maximum height.

To Find:

• The height of the tower?

Solution:

★ Finding initial velocity of ball ::

We know that,

⚘ First eqⁿ of motion,

$\bf{\dag}\:{\boxed{\underline{\underline{\bf{\pink{v = u + at}}}}}}$

Where,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time taken

We have,

• v = 0m/s

(As a ball reached it's maximum height)

• a = Acceleration due to gravity (g) = -10m/s²

(Sign is negative because acceleration is opposite to the motion of ball)

• t = 0s + 2s + 3s = 5s
• u = ?

$\tt \longrightarrow\:v = u + gt$

By putting all values we get,

$\tt \longrightarrow\:0 = u + (-10)(5)$

$\tt \longrightarrow\:0 - u = -10\:\times\:5$

$\tt \longrightarrow\:\cancel{-} u = \cancel{-} 50$

$\longrightarrow\:{\underline{\boxed{\bf{u = 50m/s}}}}\:\bigstar$

∴ Initial velocity of ball is 50m/s.

• Now let's find it's final velocity (v) of ball when it is at top of tower and time (t) = 3s.

$\tt \longrightarrow\:v = u + gt$

By putting all values we get,

$\tt \longrightarrow\:v = 50 + (-10)(3)$

$\tt \longrightarrow\:v = 50 + (-10\:\times\:3)$

$\tt \longrightarrow\:v = 50 + (-30)$

$\tt \longrightarrow\:v = 50 - 30$

$\longrightarrow\:{\underline{\boxed{\bf{v = 20m/s}}}}\:\bigstar$

∴ Final velocity of ball when it is at top of tower is 20m/s.

★ Finding the height of the tower ::

We know that,

⚘ Third eqⁿ of motion,

$\bf{\dag}\:{\boxed{\underline{\underline{\bf{\purple{v^2 = u^2 + 2as}}}}}}$

Where,

• v denotes final velocity
• final velocityu denotes initial velocity
• initial velocitya denotes acceleration

We have,

• v = 20m/s
• u = 50m/s
• a = Acceleration due to gravity (g) = -10m/s²

(Sign is negative because acceleration is opposite to the motion of ball)

• s = the height of the tower (h) = ?

$\tt \longrightarrow\:v^2 = u^2 + 2gh$

By putting all values we get,

$\tt \longrightarrow\:(20)^2 = (50)^2 + 2(-10)h$

$\tt \longrightarrow\:400 = 2500 + (2\:\times\:-10)h$

$\tt \longrightarrow\:400 - 2500 = -20h$

$\tt \longrightarrow\:\cancel{-} 2100 = \cancel{-} 20h$

$\tt \longrightarrow\:2100 = 20h$

$\tt \longrightarrow\:h = \dfrac{210\cancel{0}}{2\cancel{0}}$

$\tt \longrightarrow\:h = {\cancel{\dfrac{210}{2}}}$

$\longrightarrow\:{\underline{\boxed{\bf{h = 105m}}}}\:\bigstar$

∴ The height of the tower is 105m.

Additional Information:

✧ Three equations of motion ✧

• v = u + at
• s = ut + ½ at²
• v² = u² + 2as

✧ Some important definitions ✧

• Acceleration

Acceleration is the process where velocity changes. Since, velocity is the speed and it has some direction. So, change in velocity is considered as acceleration.

• Initial velocity

Initial velocity is the velocity of the object before the effect of acceleration.

• Final velocity

After the effect of acceleration, velocity of the object changes. The new velocity gained by the object is known as final velocity.

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