Solution Found!
A basketball is shot from an initial height of 2.40m (Fig.
Chapter 3, Problem 71GP(choose chapter or problem)
A basketball is shot from an initial height of (Fig.
) with an initial speed \(v_{0}=12 \mathrm{\ m} / \mathrm{s}\) directed at an angle \(\theta_{0}=35^{\circ}\) above the horizontal. (
) How far from the basket was the player if he made a basket? (
) At what angle to the horizontal did the ball enter the basket?
Equation Transcription:
Text Transcription:
v_0=12 m/s
theta 0=35^o
v0=12 m/s
35^o
10 ft=3.05 m
x=?
Questions & Answers
QUESTION:
A basketball is shot from an initial height of (Fig.
) with an initial speed \(v_{0}=12 \mathrm{\ m} / \mathrm{s}\) directed at an angle \(\theta_{0}=35^{\circ}\) above the horizontal. (
) How far from the basket was the player if he made a basket? (
) At what angle to the horizontal did the ball enter the basket?
Equation Transcription:
Text Transcription:
v_0=12 m/s
theta 0=35^o
v0=12 m/s
35^o
10 ft=3.05 m
x=?
ANSWER:
Step 1 of 6
A basketball is shot from an initial height of 
The kinematic equation that relates the vertical displacement of basketball and its time of flight is,
Here, 
