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.

(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.

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