Question

A ball is thrown vertically upwards with a speed of 20m/s from a tower of height 40 m. Find the maximum height from the ground covered by the ball and total time taken to reach the ground after release from the tower.

Answer 1

Answer:

Maximum height of the ball from ground = 40 meter

total time taken to reach ground

= 2(1 + √2) seconds

Step by step explanations :

Given that,

A ball is thrown vertically upwards with a speed of 20m/s from a tower of height 40 m

here,

initial velocity of the ball = 20 m/s

height of the tower = 40 m

let the height of its maximum height reached by ball from tower be h and

from ground be H

so,

H = h + 40

now,

at the maximum height velocity will 0

so,

now we have ,

Initial velocity(u) = 20 m/s

final velocity(v) = 0 m/s

height = h

gravitational acceleration(g) = -10 m/s²

[upward]

so,

by the gravitational equation of motion

v² = u² + 2gh

putting the values,

0² = (20)² + 2(-10)h

-20h = -400

h = -400/-20

h = 20 m

and we have,

H = h + 20

where,

H is maximum height of the ball from ground

so,

H = 20 + 20

H = 40 m

so,

Maximum height of the ball from ground = 40 meter

_____________________

Now,

to find time taken,

we know that,

time taken by the ball to reach its height after throwing it

= u/g

where,

u is the initial velocity

so,

time taken by ball to reach ground

= time taken (to reach its maximum height + time taken to reach ground from its maximum height)

so,

time taken to reach its maximum height

= 20/10

= 2 seconds

and we know that,

speed of ball at it's maximum height= 0

so,

u = 0 m/s

g = 10 m/s [downward]

H = 40 m

by the gravitational equation of motion

H = ut + ½ gt²

putting the values

40 = 0(t) + ½ × 10t²

5t² = 40

t² = 40/5

t² = 8

t = 2√2 s

so,

total time taken to reach ground

= 2s + 2√2s

= 2(1 + √2) seconds

_____________________

Maximum height of the ball from ground

= 40 meter

total time taken to reach ground

= 2(1 + √2) seconds

Answer 2

Solution:

Given:

A ball is thrown vertically upwards with a speed of 20 m/s from a tower of height 40 m.

Here,

$\bullet$ Initial velocity of the ball = 20 m/s

$\bullet$ Height of the tower = 40 m

Now:

Let the height reached by the ball from the tower be h and from the ground be H.

So,

$\implies$ H = h + 40

Now,

At the maximum height velocity: 0

We know that:

$\bullet$ Initial velocity (u) = 20 m/s

$\bullet$ Final velocity (v) = 0 m/s

$\bullet$ Gravitational acceleration (g) = -10 m/s²

By the gravitational equation of motion:

$\implies$ $\boxed{\sf{v^{2} = u^{2} + 2gh}}$

Substituting the values,

$\implies$ 0² = (20)² + 2(-10)h

$\implies$ -20h = -400

$\implies$ h = -400/-20

$\implies$ h = 20 m

So:

$\implies$ H = h + 20

$\implies$ H = 20 + 20

$\implies$ H = 40 m

And:

Maximum height (H): 40 m

In order to find the time taken,

We know that,

Time taken by the ball to reach its height after throwing it: $\boxed{\sf{\frac{u}{g}}}$

Where, u is the initial velocity.

So,

Time taken by ball to reach ground = Time taken (To reach its maximum height + Time taken to reach the ground from its maximum height)

So,

Time taken to reach its maximum height:

$\implies$ 20/10

$\implies$ 2 seconds

And we know that,

Speed of ball at it's maximum height: 0

So,

$\bullet$ u = 0 m/s

$\bullet$ g = 10 m/s

$\bullet$ H = 40 m

By the gravitational equation of motion:

$\implies$ $\boxed{\sf{H = ut + \frac{1}{2}\:gt^{2}}}$

Substituting the values:

$\implies$ 40 = 0(t) + ½ × 10t²

$\implies$ 5t² = 40

$\implies$ t² = 40/5

$\implies$ t² = 8

$\implies$ t = 2√2 s

So,

Total time taken to reach the ground:

$\implies$ 2s + 2√2s

$\implies$ 2(1 + √2) seconds

_____________________

Answered by: Niki Swar, Goa❤️