Question

A body is thrown horizontally from the top of a

tower and strikes the ground after three seconds at

an angle of 45º with the horizontal. Find the height

(in meter) of the tower with which the body was

projected. (Takeg 9.8 m/s^2)​

Answer 1

Answer:

Answer of this question is 9.8 m/s^2

Answer 2

Given:-

θ = 45 degree

t = 3 sec

g = 9.8 m/s^2

To find:-

s = ?

u = ?

Solution:-

As per given condition

as we know that :-

$= T = \sqrt{\frac{2h}{g} }$

$= 3 = \sqrt{\frac{2 * h}{9.8} }$

$= 9 = \frac{2h}{9.8}$

$= h = \frac{9.8 * 9}{2}$

$= h = 44.1 m$

hence , the height of tower is 44.1 m

Thank you.........