Question

A basketball player shoots toward a basket 5.6 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

A basketball player shoots toward a basket 5.6 m away and 3.0 m above the floor, so the required initial speed is approximately 4.31 m/s.

Step-by-step explanation:

To find the initial speed of the basketball, we can break the motion into horizontal and vertical components.

The horizontal component of the motion remains constant, as the player maintains his horizontal velocity.

The vertical component, however, is affected by gravity.

The ball is released at a height of 1.8 m above the floor and needs to go through the basket 3.0 m above the floor, which means it needs to rise by 1.2 m.

The required initial speed can be found using the equation:

vf² = vi² + 2 * a * d

where

vf is the final vertical velocity (0 m/s at the highest point)

vi is the initial vertical velocity

a is the acceleration due to gravity (-9.8 m/s²)

d is the vertical distance traveled (1.2 m).

Rearranging the equation, we get:

vi = sqrt(2 * a * d)

Plugging in the values, we have:

vi = sqrt(2 * (-9.8) * 1.2) = 4.31 m/s

Therefore, the initial speed of the basketball needs to be approximately 4.31 m/s in order to go through the basket.

Answer 2

9.58 m/s

Step-by-step explanation:

x = 5.6 m

y = 3 m

θ = 60°

Let u be the initial velocity of projection.

The general equation of projectile path is given by

`y = x tan\theta -(gx^(2))/(2u^(2)Cos^(2)\theta )`

`3 = 5.6* tan60 -(9.8* 5.6 * 5.6)/(2* u^(2)Cos^(2)60)`

`(614.66)/(u^(2))=6.7`

u = 9.58 m/s

thus, the velocity of projection is given by 9.58 m/s.