3.a ball is thrown into the air with an initial velocity of 15 m/s. a. how long does it take the ball to reach maximum height? b. what is the maximum height that the ball reaches? c. what is the ball’s velocity at its maximum height? d. how long does it take for the ball to return to its initial height?

 a) = 1.5 sec.  b= 11.5 m. c=0  d= 1.5 sec.

Explanation:

a) Once thrown into the air , the only influence on the ball is gravity, so, we can conclude that the acceleration acting on the ball is the one due to gravity, i.e., g= 9.8 m/sec².

Applying the definition of acceleration, we can write the following equation:

a= (vf-vo) / t   ⇒ g = vf-vo / t. (1)

Now, when the ball reaches to his maximum height, it stops momentarily, and then starts to fall, so for this height, we can write vf=0.

Replacing  in (1), and solving for t, we get:

t = (0 -vo) / g = -15 m/s/ - 9.8 m/sec² (Taking upward as positive  direction)

a) t= 1.5 sec.

Replacing this value in the kinematic equation for displacement, when vo=15 m/s, we get:

hmax: 15 m/sec . 1.5 sec -1/2. 9.8 m/sec². (1.5 sec) ² (2)

b) hmax= 11.5 m

c) vf= 0 (as explained above)

Finally, in order to know how long the ball takes to return to its initial height, we know that for the falling part of the trajectory, vo=0, so we can write the equation for displacement as follows:

h=-hmax (because it returns to the same starting point)= 1/2 gt²=-11.5 m

Solving for t, we get:

d) t=1.5 sec.

It can be seen that the time used in going up, it is the same for going down, to travel the same distance.

This result is completely general and applies to any free fall problem.

i don't think so. although it would be nice. i am pretty sure that was an option an update or two ago. i guess they want you to stay 14 forever. haha.

inventory equipment is 1st