Question

a man throws a ball vertically upwards with a velocity of 20 m /s-2 . After what time will the ball come back to his hands
plz reply fast​

Answer 1

Answer:

2 seconds

Explanation:

In the attached photo:

Answer 2

Answer:-

$\red{\bigstar}$ The time after which the ball will come back to his hands is $\bf\purple{4 \: seconds}$.

• Given:-

Velocity of the ball [u] = 20 m/s-²

• To Find:-

Time after which the ball come back to hands = ?

• Solution:-

The final velocity of the ball as it hits the thrower's hands and comes to rest then, will be 0.

$\pink{\bigstar}$$\boxed{\sf{V_{final} = U_{inital} + gt}}$

here,

v = 0 m/s

u = 20 m/s

g = - 10 m/s² [moving against the gravity]

t = ?

→ 0 m/s = 20 m/s + (-10 m/s² ) × t

→ 0 m/s = 20 m/s - 10 m/s² × t

→ 0 m/s - 20 m/s = -10 m/s² × t

→ t × -10 m/s² = -20 m/s

→ t = $\sf{\dfrac{-20 m/s}{-10 m/s^2}}$

→ t = 2 sec.

Therefore, the time taken by the ball to reach the maximum height is 2 seconds.

Also,

$\pink{\bigstar}$$\boxed{\sf{Time \: of \: ascent = Time \: of \: descent}}$

Hence,

• Time taken by the ball to reach the throwers hand will be equal to the time taken by the ball to reach it's maximum height.

Therefore,

→ 2 + 2 sec.

→ $\bf\red{4 \: seconds}$