Question

3. A ball is thrown vertically upward with a velocity of 40m/s Calculate the time taken by the ball to come back.
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Answer 1

Answer:

8 seconds

Explanation:

Given :

• Initial velocity = u = 40 m/s

• Acceleration = a = - 10 m/s²

• Final velocity = v = 0 m/s

To find :

• Time taken by the ball to come back

Using the first equation of motion :

V=u+at

0=40-10t

t = 40/10

t = 4 seconds

Using second equation of motion :

S=ut+½×at²

S=40×4+½×-10×16

S=160-80

S=80 metres

For downward motion:

• Acceleration = a = 10 m/s²

• Initial velocity = u = 0 m/s

• Distance = s = 80 metres

Using second equation of motion :

S=ut+½×at²

80=½×10×t²

80=5t²

80/5=t²

16=t²

t=4 seconds

Total time = 4+4 = 8 seconds

Answer 2

Given :-

A ball is thrown vertically upward with a velocity of 40 m/s .

Required to find :-

• Time taken by the ball to come back ?

Equations used :-

$\huge \boxed{ \text{\pink{s = ut + \frac{1}{2} a {t}^{2} }}}$

$\huge \boxed{\text{\pink{ v = u + at }}}$

Solution :-

Given information :-

A ball is thrown vertically upward with a velocity of 40 m/s

We need to find the time taken by the ball to come back !

So,

From the given information we can conclude that ;

• Initial velocity of the ball = 40 m/s

• Acceleration = - 10 m/s²

( Here, acceleration is due to the gravity . The reason why the acceleration is taken in negative is the acceleration is opposite to the direction of the motion )

• Final velocity of the ball = 0 m/s

So,

Using the 1st equation of motion ;

v = u + at

t = v - u / a

So,

t = 0 - 40 / - 10

t = - 40/ - 10

negative sign ( - ) gets cancelled in both numerator and denominator

t = 40 / 10

t = 4 seconds

Hence,

Time taken by the ball for the upward motion is 4 seconds

Similarly,

we need to find the displacement of the ball in the downward motion

So,

Using the 2nd equation of motion ,

$\large \boxed{ \text{\pink{s = ut + \frac{1}{2} a {t}^{2} }}}$

So,

$\tt \: s \: = \: 40 \times 4 + \dfrac{1}{2} \times - 10 \times 4 \times 4$

$\tt \: s \: = \: 160 + (- 5) \times 16$

$\tt s = 160 - 5 \times 16$

$\tt s = 160 - 80$

$\tt s = 80$

Hence,

displacement of the ball in upward motion = 80 meters

Now,

We need to find the time taken by the ball for the downward motion using the some of the above derived values .

So,

• Initial velocity = 0 m/s

• Acceleration = 10 m/s²

( here acceleration is positive because it is in the direction of the motion )

• Displacement = 80 meters

Using the 2nd equation of motion ;

$\tt 80 = 0 \times t + \dfrac{1}{2} \times 10 \times {t}^{2}$

$\tt 80 = 5 \times {t}^{2}$

$\implies \tt \: 5 {t}^{2} = 80$

$\tt {t}^{2} = \dfrac{80}{5}$

$\tt {t}^{2} = 16$

$\tt t = \sqrt{16}$

$\tt \: t = + 4 \: \tt \: or \: - 4$

Since time can't be in negative .

So,

Time taken by the ball in the downward motion = 4 seconds

Hence,

Time taken by the ball to come back = 4 + 4

=> 8 seconds

Therefore,

Time taken by the ball to come back = 8 seconds