Question

A baseball is hit straight up and is caught by the catcher 2.0 s later. The maximum height of the ball during
this interval is

Answer 1

Given:-

• Final Velocity = 0m/s

• Time taken = 2s

• Acceleration due to gravity = 10m/s²

To Find:-

• The Maximum height attained by ball during this interval of time.

Formulae used:-

• v = u + at
• v² - u² = 2as

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time
• s = Distance

How to solve ?

• First we have to find the Initial Velocity with which it was thrown

• Then we will use third Equation of motion

Now,

→ v = u + at

→ (0) = u + -10 × 2

→ -u = -20

→ u = 20m/s

So, The initial Velocity of baseball is 20m/s

Therefore,

→ v² - u ² = 2as

→ (0)² - ( 20 )² = 2 × -10 × s

→ -400 = -20s.

→ s = 400/20

→ s = 20m.

The maximum Distance covered by baseball is 20m.

Answer 2

Question:

A baseball is hit straight up and is caught by the catcher 2.0s later. The maximum height of the ball during this interval is?

Given:

Final velocity (v) = 0m/s

Time taken (t) = 2s

Acceleration due to gravity (a)

= 10m/s^2(assuming)

To find:

Maximum height at which it can get to in that time.

Solution:

In order to find maximum height which it can get to in that timing we need to find initial velocity first.

We know that,

v = u + at

➝ 0 = u + at

➝ 0 = u + 10 × 2

➝ u = 20 m/s

Thus, initial velocity = 20 m/s

Now by using third equation of motion

$v {}^{2} - u {}^{2} = 2as \\ \\ = > 0 {}^{2} - (20) {}^{2} = 2 \times - 10 \times s$

➝ – 400 = – 20s

➝ s = 20m

Hence, maximum height of the baseball is 20m(approx).

The time it takes the ball to reach it's maximum height is the same time it needs to return to the initial point. This is a free fall problem.