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

Answer:

  v₀ = 6.64 m / s

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 = $\frac{ v_o \ x \ sin \theta }{ v_o \ cos \theta} - \frac{1}{2} g \ \frac{x^2 }{v_o^2 \ cos^2 \theta }$

        y - y₀ = x tan θ - $\frac{ g \ x^2 }{ 2 \ cos^2 \theta } \ \frac{1}{v_o^2 }$

         

we substitute the values

        3 -1.8 = 3.0 tan 60 - $\frac{ 9.8 \ 3^2 }{2 \ cos^2 60 } \ \frac{1}{v_o^2}$

        1.2 = 5.196 - 176.4 1 / v₀²

        176.4 1 / v₀² = 3.996

        v₀ = $\sqrt{ \frac{ 176.4}{3.996} }$

        v₀ = 6.64 m / s