Answer :
Answer:
[tex]v=29.3283\ m.s^{-1}[/tex]
Explanation:
Given:
- height of point from where the ball is projected, [tex]h=12\ m[/tex]
- angle of projection of the ball below the horizontal, [tex]\theta=45^{\circ}[/tex]
- initial velocity of projection, [tex]u=25\ m.s^{-1}[/tex]
Now we find the vertical component of the initial velocity downwards:
[tex]u_y=u.\sin\theta[/tex]
[tex]u_y=25\times \cos45^{\circ}[/tex]
[tex]u_y=17.6777\ m.s^{-1}[/tex]
Now the final vertical velocity of the ball when it hits the ground:
using the equation of motion,
[tex]v_y^2=u_y^2+2.g.h[/tex]
[tex]v_y^2=17.6777^2+2\times 9.8\times 12[/tex]
[tex]v_y=23.402\ m.s^{-1}[/tex]
Since the horizontal component of the motion is uniform since no force acts in the horizontal direction:
[tex]v_x=u_x=25\cos45[/tex]
[tex]v_x=17.6777\ m.s^{-1}[/tex]
Now the resultant final velocity:
[tex]v=\sqrt{(17.6777)^2+(23.402)^2}[/tex]
[tex]v=29.3283\ m.s^{-1}[/tex]