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

Science

Class 9

NCERT

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A stone is allowed to fall from the top of a tower

100

m high and at the same time another stone is projected vertically upwards from the ground with a velocity of

25 m / s

. Calculate when and where the two stones will meet.

A ball thrown up vertically returns to the thrower after

6

s. Find

(a)

The velocity with which it was thrown up,

(b)

The maximum height it reaches, and

(c)

Its position after

4

In what direction does the buoyant force on an object immersed in a liquid act?

View all question

Question

Easy

Solving time: 4 mins

A ball thrown up vertically returns to the thrower after $6$ s. Find
$(a)$ The velocity with which it was thrown up,
$(b)$ The maximum height it reaches, and
$(c)$ Its position after $4$ s.

Views: 6,347 students

Updated on: Sep 27, 2023

Book:

Science (NCERT)

Found 4 tutors discussing this question

Sebastian Discussed

A ball thrown up vertically returns to the thrower after

6

s. Find

(a)

The velocity with which it was thrown up,

(b)

The maximum height it reaches, and

(c)

Its position after

4

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

The ball returns to the ground after

6

seconds.

Thus the time taken by the ball to reach to the maximum height

(h)

is 3 seconds i.e

t = 3

Let the velocity with which it is thrown up be

u

(a)

. For upward motion,

v = u + a t

∴

0 = u + (- 10) \times 3

⟹ u = 30

m / s

(b)

. The maximum height reached by the ball

h = u t + \frac{1}{2} a t^{2}

h = 30 \times 3 + \frac{1}{2} (- 10) \times 3^{2}

h = 45 m

(c)

. After

3

second, it starts to fall down.

Let the distance by which it fall in

1

s be

d

d = 0 + \frac{1}{2} a t^{'2}

where

t^{'} = 1

d = \frac{1}{2} \times 10 \times (1)^{2}

= 5 m

∴

Its height above the ground,

h^{'} = 45 - 5 = 40 m

Hence after

4 s

, the ball is at a height of

40 m

above the ground.

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Practice more questions from Science (NCERT)

A stone is allowed to fall from the top of a tower

100

m high and at the same time another stone is projected vertically upwards from the ground with a velocity of

25 m / s

. Calculate when and where the two stones will meet.

A ball thrown up vertically returns to the thrower after

6

s. Find

(a)

The velocity with which it was thrown up,

(b)

The maximum height it reaches, and

(c)

Its position after

4

In what direction does the buoyant force on an object immersed in a liquid act?

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Question Text	A ball thrown up vertically returns to the thrower after $6$ s. Find $(a)$ The velocity with which it was thrown up, $(b)$ The maximum height it reaches, and $(c)$ Its position after $4$ s.
Updated On	Sep 27, 2023
Topic	Gravitation
Subject	Science
Class	Class 9
Answer Type	Text solution:1 Video solution: 29
Upvotes	2910
Avg. Video Duration	10 min

A ball thrown up vertically returns to the thrower after $6$ s. Find
$(a)$ The velocity with which it was thrown up,
$(b)$ The maximum height it reaches, and
$(c)$ Its position after $4$ s.

Text solutionVerified

Filo tutor solutions (29)

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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.

Text solutionVerified

Filo tutor solutions (29)

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A ball thrown up vertically returns to the thrower after $6$ s. Find
$(a)$ The velocity with which it was thrown up,
$(b)$ The maximum height it reaches, and
$(c)$ Its position after $4$ s.