Question #9b3c7

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Question #9b3c7

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Answer 1 · 2016-07-30T16:21:24

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

$u \to Velocity of projection = 14.8 \frac{m}{s}$ $u->"Velocity of projection"=14.8m/s$

$H \to Maximum heihgt = 5.30 m$ $H->"Maximum heihgt"=5.30m$

Let

$θ \to Angle of projection above the horizontal$ $theta->"Angle of projection above the horizontal"$

$T \to Time of flight$ $T->"Time of flight"$

Now

$u cos θ \to Horizontal component of velocity$ $ucostheta->"Horizontal component of velocity"$

$u sin θ \to Vertical component of velocity$ $usintheta->"Vertical component of velocity"$

At maximum height vertcal component becomes zero

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

$\Rightarrow {sin}^{2} θ = \frac{2 g H}{u^{2}} = \frac{2 \cdot 9.8 \cdot 5.3}{{14.8}^{2}}$ $=>sin^2theta=(2gH)/u^2=(2*9.8*5.3)/14.8^2$
$\Rightarrow sin θ = \frac{\sqrt{2 \cdot 9.8 \cdot 5.3}}{14.8}$ $=>sintheta=sqrt(2*9.8*5.3)/14.8$
$sin θ = 0.69$ $sintheta=0.69$
(a) $θ = {43.5}^{\circ}$ $" "theta =43.5^@$

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

$0 = u sin θ \cdot T - \frac{1}{2} \cdot g \cdot T^{2}$ $0=usintheta*T-1/2*g*T^2$

$T = \frac{2 \cdot u \cdot sin θ}{g} = \frac{2 \cdot 14.8 \cdot 0.69}{9.8} s \approx 2.08 s$ $T=(2*u*sintheta)/g=(2*14.8*0.69)/9.8s~~2.08s$

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

$= u c o cos s θ \times \frac{T}{2} = 14.8 \cdot {cos 43.5}^{\circ} \cdot 1.04 = 11.16 m$ $=ucocossthetaxxT/2=14.8*cos43.5^@*1.04=11.16m$

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

$= u cos θ = 14.8 \times {cos 43.5}^{\circ} = 10.74 \frac{m}{s}$ $=ucostheta=14.8xxcos43.5^@=10.74m/s$

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Question #9b3c7

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