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.

Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey.

Hence, it has taken 3 s to attain the maximum height.

Final velocity of the ball at the maximum height, v = 0 m/s

Acceleration due to gravity, g = −9.8 ms−2

Using equation of motion, v = u + at, we have

0 = u + (−9.8 × 3)

⇒ u = 9.8 × 3 = 29.4 m/s

Hence, the ball was thrown upwards with a velocity of 29.4 m/s.

(b) Let the maximum height attained by the ball be h.

Initial velocity during the upward journey, u = 29.4 m/s

Final velocity, v = 0 m/s

Acceleration due to gravity, g = −9.8 ms−2

Using the equation of motion,

𝑠 = 𝑢𝑡 +{1\over2}𝑎𝑡2

ℎ = 29.4 × 3 − {1\over2} × 9.8 × 32  ⇒ ℎ = 44.1 m

Hence, the maximum height is 44.1 m.

(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.

In this case,

Initial velocity, u = 0 m/s

Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s − 3 s = 1 s.

Using the equation of motion, 𝑠 = 𝑢𝑡 + {1\over2}𝑎𝑡2

𝑠 = 0 × 1 +{1\over2} × 9.8 × 12 ⇒ 𝑠 = 4.9 m

Now, total height = 44.1 m

This means, the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.

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