Question
Thu May 24, 2012 By: Ankita Gupta

# a ball thrown vertically upwards with a speed of 19.6meter per second from the top of a tower returns to the earth in 6 sec. find the height of tower

Fri May 25, 2012
Let 'u' be the initial velocity, 'v' the final velocity, g the gravity, and h the height of the stone thrown in upward direction
.

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 is the height of tower
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