Question

A ball is thrown vertically upwards. When it reaches half of its maximum height,
it has a velocity of 7$ms^{-1}$ . What is the maximum height attained by the ball?
(g = 9.8 ms–2).

Answer 1

Answer:

$\boxed{\mathfrak{Maximum \ height \ attained \ by \ the \ ball \ (H_{max}) = 5 \ m}}$

Given:

Velocity of ball when it reaches half of its maximum height = 7 m/s

Accelration due to gravity (g) = 9.8 m/s²

Explanation:

For journey B-C; (From the figure attached)

$\rm \implies {v_f}^{2} - {v_B}^{2} = \cancel{2}(-g) \dfrac{H_{max}}{ \cancel{2}} \\ \\ \rm \implies {0}^{2} - 7 ^{2} = - 9.8H_{max} \\ \\ \rm \implies \cancel{ - }49 = \cancel{ - }9.8H_{max}\\ \\ \rm \implies H_{max} = \frac{ 49}{9.8} \\ \\ \rm \implies H_{max} = 5 \: m$

Formula used:

$\boxed{\bold{v^2 - u^2 = 2gh}}$

Answer 2

Correct question:-

A ball is thrown vertically upwards. When it reaches half of its maximum height,it has a velocity of 7m/s.What is the maximum height attained by the ball.

Given:-

→Initial velocity of the body when it has reached

half of its maximum height=7m/s

→Acceleration due to gravity=9.8m/s²

To find:-

→Maximum height attained by the ball

Solution:-

In this problem:-

•Acceleration due to gravity will be

-9.8m/s²,as the ball is going against gravity.

•If the maximum height attained is h,then half of the maximum height is h/2.

•Final velocity of the ball at the highest point will be zero,thus v=0.

Now,by using the 3rd equation of motion,we get:-

=>v²-u²=2gh

=>0-(7)²=2(-9.8)h/2

=>-49= -9.8h

=>h = -49/-9.8

=>h=5m

Thus,maximum height attained by

the ball is 5m.