Question

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Answer 1

Question :

When a cricket ball is thrown vertically upward it reaches a maximum height of 5 metres. What was the initial speed of the ball and how much time is taken by the ball to reach the highest point?

Solution :

Maximum height attained by the ball is 5 m.

❖ For a body thrown vertically upward, g is taken negative as it acts in opposite direction.

Since acceleration is constant, we can apply equation of kinematics to solve this question$.$

♦ Initial velocity of the ball :

➙ v² - u² = 2gH

➙ 0² - u² = 2(-10)(5)

➙ -u² = -100

➙ u = √100

➙ u = 10 m/s

♦ Time of ascent :

➠ v = u + gt

➠ 0 = 10 + (-10)t

➠ t = 10/10

➠ t = 1 s

Answer 2

Given

• Maximum height = 5m
• final velocity= 0
• acceleration (gravity) = 10m/sec²

To Find

• initial velocity =?
• time taken by the ball to reach the highest point =?

Concept

❖ When a ball is thrown vertically upward , it has some initial velocity and its travels the Maximum height till the final velocity becomes zero. The time taken to reach the highest point is called the time of ascent.

Solution

◕According to 3rd equation of motion

${ \underline{ \boxed{ \sf{ \color{grey}{ {v}^{2} = {u}^{2} + 2as}}}}}$

where

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes the distance

subsitute the values

$\sf{ \implies0 = {u}^{2} + 2 \times ( - 10) \times 5} \\ \sf{ \implies {u}^{2} = 100 } \\ \sf{ \implies \:u = \sqrt{100} } \\ \sf{ \implies \: u = 10m/sec}$

So the initial velocity of the ball is 10 m/sec.

Now the time of ascent :-

According to 1st equation of Motion :-

${ \underline{ \boxed { \sf{ \color{grey}v = u + at}}}}$

• here t denotes the time of ascent.

substitute the values

$\sf{ \implies0 = 10 + ( - 10) \times \: t } \\ \sf{ \implies \: - 10 = - 10t} \\ \sf{ \implies \:t = \frac{ - 10}{ - 10} } \\ \sf{ \implies \: t = 1sec}$

So the time taken by the ball to reach the highest point is 1 sec

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