Question

A basketball player shoots toward a basket 4.9 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor at an angle of 60 o above the horizontal, what must the initial speed be if it were to go through the basket

Answer 1

v₀ = 6.64 m / s

Step-by-step explanation:

This is a projectile throwing exercise

x = v₀ₓ t

y = y₀ + v_{oy} t - ½ g t²

In this case they indicate that y₀ = 1.8 m and the point of the basket is x=4.9m y = 3.0 m

the time to reach the basket is

t = x / v₀ₓ

we substitute

y- y₀ =
`( v_o \ x \ sin \theta )/( v_o \ cos \theta) - (1)/(2) g \ (x^2 )/(v_o^2 \ cos^2 \theta )`

y - y₀ = x tan θ -
`( g \ x^2 )/( 2 \ cos^2 \theta ) \ (1)/(v_o^2 )`

we substitute the values

3 -1.8 = 3.0 tan 60 -
`( 9.8 \ 3^2 )/(2 \ cos^2 60 ) \ (1)/(v_o^2)`

1.2 = 5.196 - 176.4 1 / v₀²

176.4 1 / v₀² = 3.996

v₀ =
`\sqrt{ ( 176.4)/(3.996) }`

v₀ = 6.64 m / s