Question

A basketball player is 4.22 m from

Answer 1

Answer:

3.10 m

Explanation:

Given;

Initial speed of ball u = 9.20 m/s

Angle θ = 69°

Horizontal distance from goal d = 4.22m

Resolving the initial velocity into horizontal and vertical components;

The horizontal component of the initial velocity;

uh = ucosθ

Substituting the given values;

uh = 9.2cos69°

uh = 3.30 m/s

The time taken for it to cover the horizontal distance of 4.22 m (to reach the goal);

Time = distance/speed = d/uh

time = 4.22/3.30 =. 1.279s

The time taken to reach the goal is 1.279 seconds.

To determine the height of the ball, we will resolve the vertical component of the initial velocity;

Vertical component of the ball velocity is;

Uv = usinθ

Uv = 9.20sin69°

Uv = 8.59 m/s

Applying the equation of motion;

Height h = ut - 0.5gt^2

Velocity u = Uv = 8.59 m/s

Time t = 1.279s

Acceleration due to gravity g = 9.8 m/s^2

Substituting the values;

h = 8.69(1.279) - 0.5(9.8×1.279^2)

Height h = 3.0988891 = 3.10 m

The height of ball above the release point is 3.10m

Answer 2

Answer: The height above the release point is 2.96 meters.

Explanation:

The acceleration of the ball is the gravitational acceleration in the y axis.

A = (0, -9.8m/s^)

For the velocity we can integrate over time and get:

V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

4.22m = 9.20*cos(69°)*t

t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h =  -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.