Question #b5dd9

2 Answers
Jan 28, 2018

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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#

Jan 28, 2018

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]}