Question

Assume: A 78 g basketball is launched at an angle of 45.3 ◦ and a distance of 19.6 m from the basketball goal. The ball is released at the same height (ten feet) as the basketball goal’s height. A basketball player tries to make a long jump-shot as described above. The acceleration of gravity is 9.8 m/s 2 . What speed must the player give the ball

Answer 1

Answer:

Initial speed = v₁ = 13.86 m/s

Explanation:

This problem is related to the projectile motion. Projectile motion is a two dimensional motion having horizontal and vertical component. Horizontal component of the velocity remain constant throughout his motion while vertical component vary with time.  

Given data :

Mass = m = 78 g

Angle = θ = 45.3°

Range = R = 19.6 m

Gravitational acceleration = g = 9.8 m/s²

Initial velocity = v₁ = ?

Using the formula of range of the projectile

$R = \frac{v_{1}^{2}}{g}sin2$θ

$v_{1} ^{2} = \frac{Rg}{sin2*45.3}$

$v_{1} ^{2} = \frac{(19.6)(9.8)}{sin90.6}$

v₁² = 192.09

v₁ = 13.86 m/s

Answer 2

Answer:

Explanation:

This question directly involves range of a projectile.

Data

R = 19.6m

Θ = 45.3°

g = 9.8 m/s²

But range of a projectile is

R =(U²Sin2Θ) / g

Rg = U²Sin2Θ

U² = (R * g) / Sin2Θ

U² = (19.6 * 9.8) / sin (2*45.3)

U² = 192.08 / Sin 90.6

U² = 192.0905

U = √(192.0905)

U =13.86m/s