Question

A basketball is shot by a player at a height of 2.0 m. The initial angle

was 53° above horizontal. At the highest point the ball was travelling 6
m/s. If he scored (the ball went through the rim that is 3.00 m above the
ground), what was the player’s horizontal distance from the basket?

Answer 1

The answer is "8.28 m".

Step-by-step explanation:

`\to u \cos 53^(\circ)=6 \ \text{therefore the horizontal velocity is constant alaways}\\\\\therefore`

`u= (6 * 5)/(3)= (30)/(3)= 10\ (m)/(s)\\\\`

In the projectfile when vertical displacement is 1 then 5 m comes at farther sides

Using

`S=ut-(1)/(2) gt^2\\\\1=u\sin 53^(\circ)* t-5t^2\\\\1=8t-5t^2\\\\5t^2-8t+1=0\\\\ t=(8\pm √(64-20))/(10) \\\\=(8\pm 2√(11))/(10) \\\\t_2=(8+2√(11))/(10) \\\\`

`Distance=u \cos 53^(\circ)* t_2\\\\`

`=6 * (8\pm 2√(11))/(10) \\\\=(24\pm 6√(11))/(5) \\\\= 8.28 \ m`