PHYSICS: A person looking at a building on top of which an antenna is mounted?
The horizontal distance between the person’s eyes and the building is 90.4 m. In part a the person is looking at the base of the antenna, and his line of sight makes an angle of 34.5o with the horizontal. In part b the person is looking at the top of the antenna, and his line of sight makes an angle of 37.1o with the horizontal. How tall is the antenna?
- anil bakshiLv 71 decade agoFavorite Answer
height of building above person's eye =H
height of antenna = h
a) tan (34.5) = H/90.4
b) tan (37.1) = (H+h)/90.4
h = 90.4 tan (37.1) - H
h = 90.4 [tan (37.1) - tan (34.5)]
h = 90.4 [0.7563 - 0.6873]
= 6.24 meter
- kulpaLv 44 years ago
Assuming you mean base of the development, no longer base of the cliff... preliminary velocity of the rock=40m/s very final velocity on the optimum ingredient=0 v=u-gt => 0=40-10t => t=4s (time taken to prevail in the optimum ingredient.) distance traveled in upward action=ut-a million/2(gt^2) => s=40x4 -5x16 => s=80m finished distance to be coated throughout downward action=80+35=115m time taken, one hundred fifteen=a million/2 x 10 x t^2 => t^2= 23 => t=root 23 => t=4.8s ie, that's going to take 4+4.8s =8.8s
- SteveLv 71 decade ago
Ha = 90.4*(tan37.1° - tan34.5°) = 6.239 m