Question

Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when it returns to the level where it left your hand? (Indicate the direction with the sign of your answer.)

Answer 1

The given question is incomplete. The complete question is as follows.

You throw a ball vertically upward, and as it leaves your hand, its speed is 26.0 m/s.

(a) How high (in m) does it rise above the level where it leaves your hand?

(b) How long (in s) does it take to reach its highest point?

(c) How long (in s) does the ball take to return to the level where it left your hand after it reaches its highest point?

(d) Assume that the upward direction is positive and the downward direction is negative. What is the ball's velocity (in m/s) when it returns to the level where it left your hand? (Indicate the direction with the sign of your answer.)

Explanation:

(a) For maximum height, the formula will be as follows.

           $v^{2} = u^{2} + 2as$

                 a = $v^{2} - 2gh$

or,                 h = $\frac{v^{2}}{2g}$

                        = $\frac{(26)^{2}}{2 \times 9.8}$

                        = $\frac{676}{19.6}$

                        = 34.5 m/s

Hence, it rises 34.5 m/s above the level where it leaves your hand.

(b) Time to reach maximum height is as follows.

            v = u + at

or,           v - gt = 0

                 t = $\frac{v}{g}$

                   = $\frac{26}{10}$

                   = 2.6 sec

Therefore, it will take 2.6 sec to reach its highest point.

(c)  Time taken by the ball to ascent is equal to the time it has taken to descent.

Therefore, time taken by the ball to return to the level where it left your hand after it reaches its highest point? is also 2.6 sec.

(d)  Speed of the ball will be 26 m/s in the downward direction. Hence, the velocity will be -26 m/s.