Question

A stone is thrown upwards from the ground with a velocity 'u'
a) What is the maximum height attained by the stone?
b) Draw the position-time graph of the stone during its journey.​

Answer 1

Answer:

Maximum height attained by the stone would be Vsquare/2g.

Explanation:

(a)We know,

Maximum height of an object thrown at an ngle with the X-axis,

$h = {v}^{2} \sin( \alpha ) \div 2g$

When a body is thrown vertically upwards, the angle is 90 degrees. So, the equation becomes:

$h = {v}^{2} \div 2g$

(b) The position-time graph would show the stone at the highest position h at half the time of travel, T.

Answer 2

Answer:

h = u²/2g

graph enclosed

Explanation:

A stone is thrown upwards from the ground with a velocity 'u'

a) What is the maximum height attained by the stone?

b) Draw the position-time graph of the stone during its journey.​

V² - U² = 2aS

U = u

V = 0 ( at max Height)

S = h  ( max height attained)

a = -g

=> 0² -u² = 2(-g)h

=> h = u²/2g

Taking an example where u = 50 m/s

the position-time graph of the stone during its journey till return

T S (Position)

0 0

1 45

2 80

3 105

4 120

5 125

6 120

7 105

8 80

9 45

10 0