Question

Ques :-

A ball thrown up vertically returns to the thrower after 6s. Find

a.) the velocity with which it was thrown up.

b.) the maximum height it reaches , and

c.) its position after 4s.

Answer 1

$\huge\underline\mathfrak\blue{Answer}$

a).initial velocity (u) = 29.4 m/s.

b). the maximum height it reaches is 44.1 metres.

c). 39.2 Metres

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$\huge\underline\mathfrak\blue{Explanation}$

Total time = 6 sec

Time of ascent = Time of descent = Total time/2 => 6/2 = 3 sec

_____________________

a). Given :

final velocity (v) = 0 m/s

acceleration due to gravity (g)= -9.8m/s²

time taken (t) = 3 sec

To find :

initial velocity ( u)

Solution :

We know that,

v = u + at [ 1st equation of motion ]

or, v = u + gt

Put the given values,

0 = u + (-9.8) × 3

0 = u - 29.4

29.4 = u

Therefore, initial velocity (u) = 29.4 m/s. ....(i)

______________________

b). Given :

u = 29.4 m/s [ from (i) ]

g = -9.8 m/s²

t = 3 sec

To find :

the maximum height it reaches.

Solution :

We know that,

S = ut + 1/2 at² [ second equation of motion ]

or, h = ut + 1/2gt²

put the given values,

h = 29.4 × 3 + 1/2 × (-9.8) × 3²

h = 88.2 - 44.1

h = 44.1 m

Therefore, the maximum height it reaches is 44.1 metres.

____________________

c). Given :

u = 29.4 m/s

t = 4 sec

g = -9.8 m/s²

To find :

The position of ball after 4 sec.

Solution :

We know that,

S = ut + 1/2at² [ second equation of motion ]

Or, h = ut + 1/2gt²

h = 29.4 × 4 + 1/2 ( -9.8)(4)²

h = 117.6 - 78.4

h = 39.2 m

____________________

Answer 2

$\huge\underline\mathfrak\green{Solution}$

•Time taken = 6/2 = 3 s.

v = 0 (at the maximum height )

a = - 9.8 m s-²

• Using, v = u + at, we get

0 = u - 9.8 × 3

u = 29.4 ms-¹

•Using, 2aS = v² - u², we get

S = v²- u²/2a

0 - 29.4 × 29.4/- 2× 9.8

44.1 m

•S = ut + 1/2at ²

= 0 + 1/2 × 9.8 × 1

= 4.9 m

Thanks!! ✨❤