Answer:
Explanation:
Given
First stone is thrown at an angle of [tex]\theta =40^{\circ}[/tex] Above horizontal
second stone is thrown at angle of [tex]40^{\circ}[/tex] below horizontal
let h be the height of building
Initial velocity is same in both case i.e. u=10 m/s
Horizontal velocity will remain same as there is no acceleration and there will only be vertical velocity change
for first case
[tex]v_y^2-v^2=2gh[/tex]
[tex]v_y=\sqrt{2gh}[/tex]
Net velocity at ground
[tex]37=\sqrt{(u\cos 40)^2+(v_y)^2}[/tex]
[tex]37^2=(10\cos 40)^2+2\times 9.8\times h[/tex]
[tex]h=\frac{1310.317}{2\times 9.8}=66.85 m[/tex]
For second case
[tex]u_x=u\cos 40[/tex]
[tex]v_y'=u\sin 40[/tex]
Let [tex]v_0[/tex] be the final vertical velocity
[tex]v_0^2-v_y'^2=2gh[/tex]
[tex]v_0=\sqrt{v_y'^2+2gh}[/tex]
[tex]v_0=\sqrt{(10\sin 40)+2\times 9.8\times 66.85}[/tex]
[tex]v_0=36.76[/tex]
Final velocity at ground
[tex]=\sqrt{v_0^2+u_x^2}[/tex]
[tex]=\sqrt{36.76^2+(10\cos 40)^2}[/tex]
[tex]=37.55 m/s[/tex]
