91Ó°ÊÓ

It is given that,

$v_0=82\mathrm{m}/\mathrm{s}$

$\mathrm{θ}={45}^0$

$h=30m$

Components of initial velocity are,

$\begin{array}{ll}{v}_{0x}=v_0& \cos \mathrm{θ}\\ v_{0y}=v_0& \sin \mathrm{θ}\end{array}$

For horizontal and vertical distance, the time of flight is the same.

At 45 degrees, the ball travels maximum range when it lands on the same level, but here the ball travels below the same level. So, the range can be increased. First, find the time taken by the ball to cover a vertical distance of 30 m and using that time, find the horizontal distance. The horizontal distance travelled by the ball can also be found by using a kinematic equation.

Formula is as follow:

$d=v_{0}t+\left(\frac{1}{2}\right)a{t}^2$ (i)

Using equation (i), the time can be found as,

$-30=v_0\sin \mathrm{θ}\left(t\right)+\left(\frac{1}{2}\right)(-9.8){\left(t\right)}^2$

$-30=82\sin \left(45\right)\left(t\right)-4.9\left(t^2\right)$

$-30=57.9827\left(t\right)-4.9\left(t^2\right)$

$4.9\left(t^2\right)-57.9827\left(t\right)-30=0$

Solving this quadratic equation, the values of t can be found as,

t =12.3298 s or -0.49 s

Now, take the positive value of time,

So, the time of flight is 12.3298 s.

$d=v_0\cos \left(45\right)\left(t\right)$

$$\begin{gathered} d=82\cos \left(45\right)\left(12.3298\right) \\ =714.91\mathrm{m} \end{gathered}$$

Therefore, horizontal distance travelled by ball is 714.91 m or 715 m. So, the ball travelled localid="1657024832180" $715-686=28.9\mathrm{m}$distance greater than original distance.

Rounding off to correct significant figures,$29\mathrm{m}$is the final answer.