Respuesta :
Answer:
Explanation:
Given
inclination [tex]\theta =30^{\circ}[/tex]
initial speed [tex]u=20\ m/s[/tex]
Point of release is 45 m above the ground
Considering stone to be a projectile, so time taken by projectile for its zero vertical displacement is
[tex]t_1=\frac{2u\sin \theta }{g}[/tex]
[tex]t_1=\frac{2\times 20\times \sin 30}{10}[/tex]
[tex]t_1=2\ s[/tex]
Now after completing zero vertical displacement , stone needs to travel another 45 m in downward direction with initial speed u=20\sin 30
[tex]h=u_yt+\frac{1}{2}a_yt^2[/tex]
where, [tex]h=height[/tex]
[tex]u_y=vertical velocity[/tex]
[tex]a_y=vertical acceleration[/tex]
[tex]t_0=time[/tex]
[tex]45=20\sin 30+\frac{1}{2}(9.8)(t_0)^2[/tex]
[tex]t_0^2=\frac{70}{9.8}[/tex]
[tex]t_0=2.64\ s[/tex]
thus total time time required is [tex]t=t_0+t_1=2.64+2=4.64\ s[/tex]
vertical velocity just before hitting
[tex]v_y=\sqrt{u_y^2+2\times a_y\times s}[/tex]
[tex]v_y=\sqrt{10^2+2\times 10\times 45}[/tex]
[tex]v_y=\sqrt{1000}=31.622\ m/s[/tex]
Horizontal velocity [tex]v_x=u\cos 30=17.32\ m/s[/tex]
Net velocity Just before hitting [tex]=\sqrt{v_x^2+v_y^2}[/tex]
[tex]=\sqrt{(17.32)^2+(31.62)^2}[/tex]
[tex]=\sqrt{1299.82}=36.05\ m/s[/tex]
