Question

A body is thrown horizontally from the top of a tower and strikes the ground after three seconds at an angle of `45^@` with the horizontal. Find the height of the tower and the speed with which the body was projected. (Take ` g = 9.8 m//s^2`)

Answer 1

As shown in the figure.

`u_(y) = 0` and `a_(y) = g = 9.8 m^(-2)` ltbr. `rArr s_(y) = u_(y)t +(1)/(2)a_(y)t^(2)`
`0 xx3 +(1)/(2) xx 9.8 xx (3)^(2) = 44.1 m`
Further, `v_(y) = u_(y) +a_(y)t = 0+(9.8)(3) = 29.4 ms^(-1)`
As the resultant velocity v makes an angle of `45^(@)` with the horizontal so,
`tan 45^(@) = (v_(y))/(v_(x))`
or `1 = (29.4)/(v_(x))` or `v_(x) = 29.4 ms^(-1)`
Therefore, the speed with which the body was projected (horizontal) is `u = v_(x) = 29.4 ms^(-1)`.