Physics Help?

Question

A girl throws a ball straight upward to her sister, who is in a window 3.60 m above her hand. The sister catches the ball in her outstretched hand 1.50 s later. ​(10 points) 

a. With what initial velocity was the ball thrown? 

b. What was the velocity of the ball just before it was caught? 

c.​Was the ball traveling upward or downward just before it was caught?

Answer 1

(a)

From the given data we can understand that the ball was thrown vertically upwards ,

So for Verticall upword moving bodies have a sufficient intial velocity (u)

We use the relation s = ut +½ at²

But for vertically upward moving objects acceleration due to gravity is negitive a= -g

h = ut - ½ gt²    

Where h is height , t is time , u is intial velocity

3.6 = u(1.50) - ½ (9.8)(1.50)²

u(1.50) = 3.6 +11.025

u(1.50) = 14.625

u = 9.75

 Initial velocity of ball when thrown u = 9.75 m/s (Upward)

Answer 2

(b)

To find the velocity of ball just before it was caught , we use the relation v =u+at

Where u is intial velocity = 9.75 m/s,

a is acceleration due to gravity nagitive in this case ( a = -g ) & t =1.5 s

v = u -gt

v = 9.75 - (9.8)(1.5)

v = 9.75 - 14.7

v =  - 4.95 m/s

velocity of the ball just before it was caught is v =  4.95 m/s (downwards)

Answer 3

(c)

From (b) we clearly notice that the velocity of ball just before it was caught is v =  - 4.95 m/s

Here the negitive sign indicate that the sister caught the ball in way back , 

Means it reaches its maximum height and in the return she caught the ball

So the ball traveling downward just before it was caught