The basketball player must throw the ball at an initial speed of approximately 23 ft/s at an angle of 55° above the horizontal to make the foul shot. This result is based on the given conditions:
- Initial height h1=7.0h_1=7.0h1=7.0 ft
- Final height h2=10.0h_2=10.0h2=10.0 ft
- Horizontal distances d1=1.0d_1=1.0d1=1.0 ft and d2=14.0d_2=14.0d2=14.0 ft, giving a net horizontal distance of 13 ft
- Gravitational acceleration g=32 ft/s2g=32,\text{ft/s}^2g=32ft/s2
Using the projectile motion formula for initial speed:
v0=xcosθg2(xtanθ−y)v_0=\frac{x}{\cos \theta}\sqrt{\frac{g}{2(x\tan \theta -y)}}v0=cosθx2(xtanθ−y)g
where
- x=13 ftx=13,\text{ft}x=13ft (horizontal distance),
- y=h2−h1=3 fty=h_2-h_1=3,\text{ft}y=h2−h1=3ft,
- θ=55∘\theta =55^\circ θ=55∘,
substituting these values yields v0≈23 ft/sv_0\approx 23,\text{ft/s}v0≈23ft/s
. This calculation accounts for the vertical displacement and horizontal range of the ball to ensure it reaches the basket at the correct height and distance.
