Question

Question #b5dd9

Answer 1

Suppose that the ball was thrown at `u m/s` to achieve the given condition.

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

So, we can use, `v^2=u^2-2gs` (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 m/s`

Answer 2

The initial velocity is `=10.64ms^-1`

Explanation:

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

Then the coordinates of the hoop are `=(10, 1.05)`

Let the initial velocity of the ball is `=Vms^-1`

The vertical component of the ball is

`y=Vsin40t-1/2g t^2`............`(1)`

And

The horizontal component is

`x=Vcos40t`.....................`(2)`

`t=x/(Vcos40)`

Substituting this value of `t` in equation `(1)`

`y=Vsin40*x/(Vcos40)-1/2g*(x/(Vcos40))^2`

`y=xtan40-(gx^2)/(2V^2cos^2 40)`

Let the acceleration due to gravity be `g=10ms^-2`

Plugging in the coordinates of the hoop `(10, 1.05)`

`1.05=10tan40-((10*100)/(2*V^2cos^2 40))`

Solving for `V` in this equation

`980/(2*V^2cos^2 40)=7.34`

`2V^2=1000/((7.34*cos^2(40)))=226.25`

`V=sqrt(227.55/2)=10.64ms^-1`

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