Answer to Question #275792 in Mechanics | Relativity for helpme

Question

A basketball player throws a desperation shot at the buzzer from the backcourt

1.0 second before the game ended with the score tied. If the ball was released at a

velocity of 12.0 m/s at an angle of 38O and if the release was made at the same level as

the basket from a distance of 14.2 m, did he score? Why?

complete answer and solution plss..help me

Answer 1

Let us estimate the time needed for the ball to reach the basket.

Let x-axis be directed from the player to the basket horizontally and y-axis be directed vertically upwards. The projection of the initial velocity on the x-axis is $V_{0,x} = V_0\cos\alpha$ and the projection on the y-axis is $V_{0,y}=V_0\sin\alpha.$

Due to the same height of the initial and final points, we obtain the time twice the time needed to reach the highest point of the trajectory. At the highest point the ball has zero vertical velocity, so the time needed to reach this point is $t=V_{0,y}/g=V_0\sin\alpha/g.$

The total time of the motion is $2t=2V_0\sin\alpha/g =2\cdot12\cdot\sin38^\circ/9.81 = 1.5\,\text{s}.$

So the ball reaches the basket too late.

Let us determine the distance traveled by the ball horizontally.

$L=V_{0,x}2t = \dfrac{V_0^2\sin(2\alpha)}{g} = 14.24\,\text{m}.$

So the distance is correct, but the time interval is too big.