Answer to Question #185362 in Mechanics | Relativity for Cool

Question

A ball is thrown at 20 m/s from the roof of a building 25 high at an angle of 30 degrees below the horizontal. At what height above the ground will the ball strike the side of another building 20 meters away from the first

Answer 1

Given

V = 20 m/s

V_x= Vcos30^o = $20×\frac{\sqrt{3}}{2} = 17.32m/s$

V_y= Vsin30^o=$20×\frac{1}{2}=10m/s$

Time of flight =$\frac{2V_y}{g}=\frac{2Vsin30}{g}=2sec$

Range=$U_x×(time \space of \space flight)=17.32×2$

$=34.64m$

Range - 20m = extra length

Extra length = $34.64-20 =14.64m$

Time for extra length

S = U_x×t

14.64=17.32×t

t=0.84seconds

Now

We know that total time of flight was 2 seconds , so

Time till hitting if wall = 2 - 0.84

= 1.16 seconds

So, flight distance from top of building is

$S = ut- \frac{1}{2}gt^2$

$S = 10×1.16- \frac{1}{2}10×(1.16)^2$

=4.9m

So , length above the ground , it must be

25 - 4.9

= 20.1 m answer