Question

A player from team A was awarded two free throws. On the first attempt, he released the ball with a velocity of 6.0m/s at 45* above the horizontal from the height of 2.0 m from the floor. (a) Is this a miss? (b) For the second attempt, the player shoots at 45* from the same point. What must be the initial speed of the ball so that it will go through the basket? The basket is 4.21 m from the free throw line and 3.05 m above the floor.
heeellppp meee pleassee...T^T​

Answer 1

Answer:

a)

Hmax=v2sin⁡2452g+hH_{max}=\frac{v^2\sin^2{45}}{2g}+hHmax​=2gv2sin245​+h

Hmax=2+62sin⁡2452(9.81)=2.92 mH_{max}=2+\frac{6^2\sin^2{45}}{2(9.81)}=2.92\ mHmax​=2+2(9.81)62sin245​=2.92 m

It is less than the height of the basket of 3.05 m. So, this is a miss.

b)

d=R=v2sin⁡(2⋅45)g=v2gd=R=\frac{v^2\sin{(2\cdot45)}}{g}=\frac{v^2}{g}d=R=gv2sin(2⋅45)​=gv2​

4.21=v29.814.21=\frac{v^2}{9.81}4.21=9.81v2​

v=6.43msv=6.43\frac{m}{s}v=6.43sm​