Mubarak jumps and shoots a field goal from the far end of the court into the basket at the other end, a distance of 27

Physics May 15, 2021

Mubarak jumps and shoots a field goal from the far end of the court into the basket at the other end, a distance of 27.6 m. The ball is given an initial velocity of 17.1 m/s at an angle of 40.0 degrees to the horizontal from a height of 2.00 m above the ground. What is its velocity as it hits the basket 3.00 m off the ground?

Expert Solution

Answer:

v_horizontal=17.1*cos40
=13.09 m/s
v_vertical=17.1*sin40
=10.99m/s
now
since no horizontal force horizontal velocity will not change
vertical dispalcement=(3-2)=1
v^2=u^2-2*a*s
=10.99^2-2*9.8*1
v_vertical =10 m/s
velocity=sqrt(10^2+13.09^2)
=16.4 m/s

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