Question

A ball is thrown upward in the air with an initial velocity 40 m per second. How long does it take to reach back to the point it was thrown from ?

Answer 1

Answer:

the positive direction of the y axis is upward, then the acceleration is  g=−9.81ms2 .

The magnitude of the velocity is  v=v0+gt=40ms−1−9.81ms−2t  so the velocity decreases until  v=0 . The time you need is just found solving the above equation when  v=0 .

For the height, there is a better solution. Initial kynetic energy is  Ek=1/2mv2o  and when the ball reaches its maximum height, all of it is transformed into potential energy  Ep=mgh .

You can find  h  just solving the equation  Ek=Ep .

Explanation:

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Answer 2

8 seconds

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Given

➺ Initial velocity of the ball (u) = 40 m/s

To find

➺ Time taken to reach ground

Solution

☯ 1st Method

Considering the motion from ground to ground, the displacement (S) is 0 . Since, the initial and final position is same

Now, using equation of motion

➺ S = ut + ½ at²

➺ 0 = 40t + ½ × (-10) t²

➺ 0 = 40t - 5t²

➺ 5t² - 40t = 0

➺ 5t (t - 8) = 0

➺ t = 0 ; t = 8

This means that the ball is in ground at t = 0s and t = 8s

☯ 2nd Method

Apply the formula

➺ Time of flight (T) = 2u/g

➺ T = (2 × 40) / 10

➺ T = 80 / 10

➺ T = 8s

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Note: Taking g = -10 m/s²