Question

a ball trown vertically upward with a speed of 19.6 m per sec from a top of a tower and returns to the earth in 6sec find the height of the tower

Answer 1

we have the vertical downward motion of ball from the top till the ground,
u= -19.6 m per second
a= 9.8 m per sec square
t= 6 sec
using, s (h)= ut +1/2 at square
h = -19.6×6+1/2×9.8×6×6
h= -117.6+ 176.4
h = 58.8 m Ans....
h=

Answer 2

.
we know that v = u - gt
by squaring on both side, we get
v^2 = (u^2) - 2ugt + (gt)^2
v^2 = (u^2) - 2g (ut -(gt^2)/2)

substitute h in the place of (ut - (gt^2)/2)/
v^2 = (u^2) - 2gh
Here final velocity is zero.
0 = (u^2) -2gh
2gh = u^2
h = (u^2)/(2g)
where g = 9.8 m^2/s
h = (u^2)/19.6 m
 h = (u^2)/19.6 meter

h = 19.6 m

 

from

v = u + at

0 = 19.6 - 9.8 * t

t = 2 second to reach from the tower to max height

 

Now, from maximum height to reach to earth , the time left is 4 seconds

 now, v = 0

S = ut + 1/2 at2

(h+ 19.6) = 0 + 1/2 * 9.8 * 4 * 4

h + 19.6 = 78.4

 

h = 78.4 - 19.6 = 58.8 m