Question

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.

Answer 1

(a)

29.4 m/s

(b)

44.1 m

(c)

39.2 m above the ground

(a) 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

Acceleration due to gravity, g = −9.8 m s⁻²

Equation of motion, v = u + gt will give,

0 = u + (−9.8 × 3)

u = 9.8 × 3 = 29.4 ms^{− 1}

Hence, the ball was thrown upwards with a velocity of 29.4 m s⁻¹.

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

Initial velocity during the upward journey, u = 29.4 m s⁻¹

Final velocity, v = 0

Acceleration due to gravity, g = −9.8 m s⁻²

From the equation of motion,

(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

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.

Equation of motion, will give,

Total height = 44.1 m

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