Question

In a basketball game, a ball leaves a player's hand 6.1 m downrange from the basket from a height of 1.2 m below the level of the basket. If the initial velocity of the ball is 7.8m [55 degrees above the horizontal] in line with the basket, will the player score a basket? If not, by how much will the ball miss the basket?

Answer 1

Answer:

The ball missed the basket by 1.6 meters

Explanation:

Projectile Motion

It's when an object is moving in a two-dimensional space, being thrown with an initial speed and angle respect to the horizontal direction. The horizontal distance traveled by the object is given by

$x=vo.cos\theta .t$

And the vertical height above the level of the launch is

$\displaystyle y=vo.sin\theta. t-\frac{gt^2}{2}$

The basketball player throws the ball with initial speed $v_o=7.8 m/s$ at an angle of $\theta=55^o$. The horizontal distance to the basket (as suggested by the 'downrange' word) is x=6.1 m. We need to find out if the ball reaches the basket at a height y=1.2 m above the launching point.

From the equation

$x=vo.cos\theta .t$

We'll solve for t and find the flight time until the horizontal distance is reached by the ball

$\displaystyle t=\frac{x}{vo.cos\theta}$

$\displaystyle t=\frac{6.1}{(7.8).cos55^o}=1.363\ sec$

Now, we use this time to find y

$\displaystyle y=(7.8).sin55^o . (1.363)-\frac{(9.8)(1.363)^2}{2}$

$y=-0.4\ m$

The negative sign indicates the ball fell below the launching point, and missed the basket by

$1.2+0.4 = 1.6\ m$