Answer to Question #82020 in Mechanics | Relativity for Tia Elmasry

Question

A ball is thrown at 33∘ below the horizontal from a rooftop of height 95m. It lands 2.6s later. Find:
the angle at which it hits the ground

Answer 1

Input Data:

Time:

t=2.6s

Height:

h=95m

Ball throw angle:

α=33°

Solution:

The tangent of the angle of flight of the ball is:

tg β=V_y/V_x

Ball motion equation:

0=h-Vtsin α-(gt^2)/2

From here we get the value of the speed of the ball at the moment of falling:

V=h/(t*sin α)-gt/(2sin α)=95/(2.6*sin (33°))-(9.8*2.6)/(2sin (33)°)=67.1-23.4=43.7m/s

X axis velocity:

V_x=Vcos α

Y axis velocity:

V_y=Vsin α+gt

tg β=(Vsin α+gt)/(Vcos α)=tg α+gt/(Vcos α)=0.65+0.7=1.35

Ball landing angle:

β=arctg(1.35)=53.5°