# Question #9b3c7

Jul 30, 2016

#### Explanation:

When the ball reaches the target at maximum height H ,the ball will have velocity only in horizontal direction and it will be done during half of the time of flight.

Given

$u \to \text{Velocity of projection} = 14.8 \frac{m}{s}$

$H \to \text{Maximum heihgt} = 5.30 m$

Let

$\theta \to \text{Angle of projection above the horizontal}$

$T \to \text{Time of flight}$

Now

$u \cos \theta \to \text{Horizontal component of velocity}$

$u \sin \theta \to \text{Vertical component of velocity}$

At maximum height vertcal component becomes zero

$0 = {u}^{2} {\sin}^{2} \theta - 2 \cdot g \cdot H$

$\implies {\sin}^{2} \theta = \frac{2 g H}{u} ^ 2 = \frac{2 \cdot 9.8 \cdot 5.3}{14.8} ^ 2$
$\implies \sin \theta = \frac{\sqrt{2 \cdot 9.8 \cdot 5.3}}{14.8}$
$\sin \theta = 0.69$
(a)$\text{ } \theta = {43.5}^{\circ}$

Again after T time the net vertical displacement will be zero. So

$0 = u \sin \theta \cdot T - \frac{1}{2} \cdot g \cdot {T}^{2}$

$T = \frac{2 \cdot u \cdot \sin \theta}{g} = \frac{2 \cdot 14.8 \cdot 0.69}{9.8} s \approx 2.08 s$

The ball will reach the target after T/2 sec of its relrase i.e.after 1.04s.
(b)So the horizintal distance to be covered to reach the target

$= u c o \cos s \theta \times \frac{T}{2} = 14.8 \cdot \cos {43.5}^{\circ} \cdot 1.04 = 11.16 m$

(c)The speed of the ball when it reaches the target is nothing but the undiminshed horizontal component of the velocity of projection.

$= u \cos \theta = 14.8 \times \cos {43.5}^{\circ} = 10.74 \frac{m}{s}$