Physic question.

Question

In the 1970 National Basketball Association championship, Jerry West made a 60 ft shot from beyond half court to lead the Los Angeles Lakers to an improbable tie at the buzzer with the New York Knicks. Suppose West threw the ball at an angle of 40∘ above the horizontal. The basket is 10ft from the court floor.

 Neglecting air resistance, estimate the initial speed 𝑣0 of the ball.

(Resolve the initial velocity into its horizontal and vertical components using trigonometry. Be sure to use the correct component in the horizontal or vertical equations of motion.

Taking into account Jerry West's height of 6 ft 2 in,6 ft 2 in, the length of his arms, and any added jump height, a reasonable approximation is that the initial height of the ball equals the final height of the ball. This approximation simplifies the algebra considerably.)

Answer 1

Set down the Equation Battery for Ballistic Trajectory:

v_x = v_x0 equal to v₀cos θ₀ which is constant;----------------Horizontal Velocity Component

v_y = v_y0 − gt equal to v₀sin θ₀ − gt;-----------------------------Vertical Velocity Component

x = v_x0t equal to (v₀cos θ₀)t;-------------------------------------Horizontal Position Component

y = v_y0t − 0.5gt² equal to (v₀sin θ₀)t − 0.5gt².----------------Vertical Position Component

Write the Horizontal Position Component as

60 ft = (v₀cos 40°)t.

Jerry West should have enough "leg-spring" and arm-length to launch the ball at the

height of the basket, so write the Vertical Position Component as

0 ft = (v₀sin 40°)t − 0.5(32.17405 ft/s²)t².

Then v₀ is equal to 60/(t cos 40°) as well as 16.087025t/sin 40°.

Obtain from this last 60sin 40° = (16.087025cos 40°t²) which forces

t to 1.76906798 seconds for time of flight from Jerry West's hand to

the basket.

Then both 60/(t cos 40°) and 16.087025t/sin 40° will give v₀ as

44.2744079 feet per second.