Question

A ball thrown vertically upwards with an initial velocity 49m/s calculate the maximum height attained by the ball

Answer 1

Answer:

Explanation:

Given parameters

Initial velocity of the ball (u) = 49m/s.

The velocity of the ball at maximum height (v) = 0.

g = 9.8m/s2

Let us considered the time taken is t to reach the maximum height H.

Consider a formula,

2gH = v2 – u2

2 × (- 9.8) × H = 0 – (49)2

– 19.6 H = – 2401

H = 122.5 m

Now consider a formula,

v = u + g × t

0 = 49 + (- 9.8) × t

– 49 = – 9.8t

t = 5 sec

(1) The maximum height to ball rises = 122.5 m

(2) The total time ball takes to return to the surface of the earth = 5 + 5 = 10 sec.

Answer 2

GiveN :

• Final velocity (v) = 0 [Velocity of the ball at maximum height]

• Initial velocity of the ball (u) = 49m/s

• Acceleration due to gravity, g = 9.8m/s²

AssumptioN :

Let us assume that the maximum height attained by the ball be 'h'.

In order to find the maximum height attained by the ball, we'll use the equations of motion.

By using the 2nd equation of motion :

$\longrightarrow\sf v^2 - u^2 = 2gh$

Substituting the values :

$\longrightarrow\sf 2 \times (- 9.8) \times h = 0 - (49)^2$

$\longrightarrow\sf -19.6h = -2401$

$\longrightarrow\sf h = \dfrac{2401}{19.6}$

$\longrightarrow \large{ \boxed{\sf {h = 122.5 \: m}}}$

$\therefore$ The maximum height to ball attained is 122.5 m.

Something that helps :

Basically, there are 3 types of equations of motion. They're :

$\bullet\sf\;\;\;\;\;v = u + at$

$\bullet\sf\;\;\;\;\;v^2 = u^2 + 2as$

$\bullet\sf\;\;\;\;\;S = ut + \dfrac{1}{2}at^2$

Also, equations of motion under gravitational force can be applied by just replacing the acceleration by acceleration due to gravity (g).