# Question b5dd9

Jan 28, 2018 Suppose that the ball was thrown at $u \frac{m}{s}$ to achieve the given condition.

Now, considering vertical motion,it has to reach $\left(3.05 - 2\right) m$ or, $1.05 m$ and this will be the maximum height,for just to cross the basket.

So, we can use, ${v}^{2} = {u}^{2} - 2 g s$ (all the symbols are bearing their conventional meaning)

Where, $v = 0 , a = g$,$u = u \sin 40$ i.e the vertic component of velocity,and, $s = 1.05$

So, u^2= 2×10×1.05/(sin 40)^2#

Or, $u = 7.15 \frac{m}{s}$

Jan 28, 2018

The initial velocity is $= 10.64 m {s}^{-} 1$

#### Explanation:

Let 's have a rectangular coordinate system $\left(x , y\right)$ with the origin at the starting point of the ball,

Then the coordinates of the hoop are $= \left(10 , 1.05\right)$

Let the initial velocity of the ball is $= V m {s}^{-} 1$

The vertical component of the ball is

$y = V \sin 40 t - \frac{1}{2} g {t}^{2}$............$\left(1\right)$

And

The horizontal component is

$x = V \cos 40 t$.....................$\left(2\right)$

$t = \frac{x}{V \cos 40}$

Substituting this value of $t$ in equation $\left(1\right)$

$y = V \sin 40 \cdot \frac{x}{V \cos 40} - \frac{1}{2} g \cdot {\left(\frac{x}{V \cos 40}\right)}^{2}$

$y = x \tan 40 - \frac{g {x}^{2}}{2 {V}^{2} {\cos}^{2} 40}$

Let the acceleration due to gravity be $g = 10 m {s}^{-} 2$

Plugging in the coordinates of the hoop $\left(10 , 1.05\right)$

$1.05 = 10 \tan 40 - \left(\frac{10 \cdot 100}{2 \cdot {V}^{2} {\cos}^{2} 40}\right)$

Solving for $V$ in this equation

$\frac{980}{2 \cdot {V}^{2} {\cos}^{2} 40} = 7.34$

$2 {V}^{2} = \frac{1000}{\left(7.34 \cdot {\cos}^{2} \left(40\right)\right)} = 226.25$

$V = \sqrt{\frac{227.55}{2}} = 10.64 m {s}^{-} 1$

graph{(y-0.839x+0.0735x^2)(y-1.05)=0 [1.31, 21.31, -1.11, 8.89]}