A man chased by the army, leaps with speed 12 m/s at a 47 degree angle above the horizontal from the top of a building. If he lands 40.0 m from the base of the building, then finds the height of the building

Answer 1

Vo = 12m/s[47].
Xo = 12*Cos47 = 8.18 m/s.
Yo = 12*sin47 = 8.78 m/s.

Xo*T = 40.
8.18*T = 40, T = 4.89 s. = Time in air.

Y = Yo + g*Tr_.
0 = 8.78 - 9.8Tr, Tr = 0.896 s. = Rise time.

Tr+Tf = 4.89.
0.896 + Tf = 4.89, Tf = 3.99 s. = Fall time.

ha = Yo*Tr + 0.5g*Tr^2.
ha = 8.78*0.896 - 4.9*0.896^2 = 3.93 m. = Ht. above the roof.

h = 0.5g*Tf^2 = 4.9*3.99^2 = 78 m. Above gnd.

ha+hb = 78 m.
3.93 + hb = 78, hb = 74.1 m. = Ht. of the bldg.