Question

3. Someone at the third-floor window (12.0 m above the ground) hurls a ball downward at an angle of

Answer 1

Answer:

$v=29.3283\ m.s^{-1}$

Explanation:

Given:

• height of point from where the ball is projected, $h=12\ m$

• angle of projection of the ball below the horizontal, $\theta=45^{\circ}$

• initial velocity of projection, $u=25\ m.s^{-1}$

Now we find the vertical component of the initial velocity downwards:

$u_y=u.\sin\theta$

$u_y=25\times \cos45^{\circ}$

$u_y=17.6777\ m.s^{-1}$

Now the final vertical velocity of the ball when it hits the ground:

using the equation of motion,

$v_y^2=u_y^2+2.g.h$

$v_y^2=17.6777^2+2\times 9.8\times 12$

$v_y=23.402\ m.s^{-1}$

Since the horizontal component of the motion is uniform since no force acts in the horizontal direction:

$v_x=u_x=25\cos45$

$v_x=17.6777\ m.s^{-1}$

Now the resultant final velocity:

$v=\sqrt{(17.6777)^2+(23.402)^2}$

$v=29.3283\ m.s^{-1}$