Question

A ball is thrown vertically upwards, with a velocity of 49m/s. Calculate the maximum height reached by the ball, time taken by the ball to reach that height, time taken by the ball to return back to the ground from the highest point​

Answer 1

Answer

1. 120.05 m
2. 4.9 s
3. 4.9 s

Given

A ball is thrown vertically upwards, with a velocity of 49 m/s

To Find

1. Maximum height reached by the ball
2. Time taken by the ball to reach that height
3. Time taken by the ball to return back to the ground from highest point

Concept Used

We need to apply equation's of kinematics .

→ v = u + at

→ s = ut + ¹/₂ at²

→ v² - u² = 2as

Solution

Initial velocity , u = 49 m/s

Final velocity , v = 0 m/s

[ ∵ After reaching to certain height , it goes to rest ]

Acceleration due to gravity , a = - g = -10 m/s²

[ ∵ thrown against the gravity ]

Height , s = ? m

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (0)² - (49)² = 2(-10)s

⇒ 0 - 2401 = - 20s

⇒ 20s = 2401

⇒ s = 120.05 m

1 .

So , maximum height reached by the ball is 120.05 m

______________________________

Apply 1 st equation of motion ,

⇒ v = u + at

⇒ (0) = (49) + (-10)t

⇒ 0 = 49 - 10t

⇒ 10t = 49

⇒ t = 4.9 s

2 .

Time taken by the ball to reach maximum height = 4.9 s

______________________________

Now time taken by the ball to return back to ground from the highest point is equals time taken by the ball to reach highest point .

⇒ t = 4.9 s

3 .

t = 4.9 s

Answer 2

✰ Given -

• Initial Velocity (u) = 49 m/s
• Final Velocity (v) = 0 m/s
• Deceleration (Gravity) = -10 m/s^2

✰ To Find -

• Max Height Reached by ball
• Time taken to reach Max Height
• Time taken to reach ground from that Max Height

✰ Solution -

We know 3rd Equation of Kinematics,

$✓ \: {v}^{2} - {u}^{2} = 2as$

Placing Values,

$⟶( {0}^{2} ) - ( {49}^{2} ) = 2 \times ( - 10) \times s$

$⟶0 - 2401 = - 20s$

$⟶20s = 2401$

$⟶s = \frac{\cancel{2401}}{\cancel{20}}$

$⟶s = 120.05m$

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We Know 1st Equation of Kinematics,

$✓v = u + at$

Placing Values,

$⟶0 = 49 + ( - 10) \times t$

$⟶0 = 49 - 10t$

$⟶10t = 49$

$⟶t = \frac{\cancel{49}}{\cancel{10}}$

$⟶t = 4.9sec$

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As we know that, Time taken for object to reach Max Height is equal to time taken to reach ground from Maximum Height!

$⟶t = 4.9sec$

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