Question

Question #9b3c7

Answer 1

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->"Velocity of projection"=14.8m/s`

`H->"Maximum heihgt"=5.30m`

Let

`theta->"Angle of projection above the horizontal"`

`T->"Time of flight"`

Now

`ucostheta->"Horizontal component of velocity"`

`usintheta->"Vertical component of velocity"`

At maximum height vertcal component becomes zero

`0=u^2sin^2theta-2*g*H`

`=>sin^2theta=(2gH)/u^2=(2*9.8*5.3)/14.8^2`
`=>sintheta=sqrt(2*9.8*5.3)/14.8`
`sintheta=0.69`
(a)`" "theta =43.5^@`

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

`0=usintheta*T-1/2*g*T^2`

`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

`=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.

`=ucostheta=14.8xxcos43.5^@=10.74m/s`