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A body is thrown horizontally from the t...

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 . Then find ,
(i) The height of the lower
(ii) The speed of projection of the body

Text Solution

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(i) Let H be the height of the tower
The time of flight `T_(f) = sqrt((2H)/(g)) = 3 `s
`rArr " H " = (g xx (3)^(2))/(2)`
= `(9.8xx9)/(2)`
= 44.1 m
(ii) Let the speed of projection be `v_(0)`
then for horizontal projection
`v_(x) = v_(0)`
`v_(y) = -"gt"`
At `" t " = T_(f) = 3 `s,
`v_(y) = 9.8 xx 3 `
`= - 29.4 m s^(-1)`
The angle which the final velocity makes with the horizontal = `theta = 45^(@)` (Given)
`rArr " " - tan 45^(@) = (-v_(y))/(v_(x))`
`rArr " " v_(y) = v_(x)`
So `v_(x) = 29.4 m s^(-1)`
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