Suppose Ball 1 is at a height of 19.6 m when it reaches the top of its trajectory. At the instant Ball 1 is at the top of its trajectory, Ball 2 is released from rest at the same height. How much time does it take for either ball to reach the ground? Note that the vertical component y of the ball’s displacement points downward, in the -y direction.

Both ball will take approximately 2 sec to reach the ground.

Explanation:

Height reached by ball 1 at the top of its trajectory = 19.6 m

Assumptions:

Using Newton's equation of motion,

s = ut + g

where s is the distance traveled down = 19.6 m

t is time taken to travel down = ?

g is acceleration due to gravity = 9.81

u is the initial velocity which is zero for both balls

substituting, we have

19.6 = (0 x t) +( x 9.81 x )

19.6 = 0 + 4.905

= 3.996

t = 1.99 sec ≅ 2 sec

multiply 2kg by latent heat of vap

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr