Question

1. (6 pts) Ball (a) On the figure above, label the positions with the highest speed (A), the lowest speed (B) and the highest velocity (C). (note, there can be more then one of some of these positions) (b) If a ball is thrown straight up at 20 m/sec, how long does it take to get back to your hand?

also wanted to figure out how high the ball went?

Answer 1

Here the ball is thrown upward with velocity 20 m/s.

Initial speed of the ball is maximum i.e 20 m/s and gradually the speed of the ball decreases till it reaches the highest point because of the acceleration due to gravity which is in opposite direction to the velocity of the ball .

When the ball reaches the highest point its velocity becomes 0 m/s and after that the ball comes downwards (because of the acceleration due to gravity).Finally when the ball reaches the point of projection , its speed will be same as the initial speed .

Thus the position where the ball has highest speed is the point of projection .Thus point A is at y = 0(see fig.)

Lowest speed of the ball becomes 0 m/s which is at the highest point .Thus B is at the highest point .(see fig.)

BT Y 4 J0 (C)

The highest velocity of the ball is at the point of projection .Thus point C is at y = 0 .

b ) .

For constant acceleration ' a '

$s=ut+\frac{1}{2}*a*t^{2}$

u = initial velocity

t = time

s = displacement

From the point of projection to back into your hand

s = 0

u = +20 m/s

a = -9.8 m/s (acceleration due to gravity)

$0=20*t+\frac{1}{2}*(-9.8)*t^{2}$

$4.9t^{2}-20t=0$

$t(4.9t-20)=0$

t = 0 , $\frac{20}{4.9}$

t = 4.08 sec

at t = 4.08 sec the ball will take to come back into the hands .