A basketball is shot from 2 meters up at an angle of 60° above the x axis at an initial velocity of 9 m/s.What is the maximum height the ball will reach?
initial distance up = 2
initial velocity component up = 9 sin 60 = 7.79
v = 9 sin 60 - 9.8 t
when v = 0, we are there
9.8 t = 7.79
t = .795 seconds to top
h = 2 + 7.79(.795) - 4.9(.795^2)
To find the maximum height the basketball will reach, we can use the principles of projectile motion. The motion of the basketball can be broken down into horizontal and vertical components.
First, let's calculate the initial vertical velocity of the basketball. The initial velocity is given as 9 m/s at an angle of 60° above the x-axis. Since the angle is measured above the x-axis, we need to find the vertical component of the velocity.
Vertical component of initial velocity (Vy) = initial velocity * sin(angle)
Vy = 9 m/s * sin(60°)
Using the value of sin(60°) = √3/2:
Vy = 9 m/s * √3/2
Vy = 4.5√3 m/s
Next, let's calculate the time taken for the basketball to reach its maximum height. At the maximum height, the vertical component of velocity becomes zero.
The equation for vertical velocity as a function of time is:
Vy = initial velocity + acceleration * time
Since the vertical velocity at the maximum height is zero, we can solve for the time taken (t) using the equation:
0 = 4.5√3 m/s - 9.8 m/s^2 * t
Simplifying the equation, we can find the value of t:
4.5√3 m/s = 9.8 m/s^2 * t
t = (4.5√3 m/s) / (9.8 m/s^2)
t ≈ 0.76 seconds
Now that we have the time taken to reach the maximum height, we can find the maximum height (h) using the equation for vertical displacement in projectile motion:
h = initial vertical velocity * time - (1/2) * acceleration * time^2
h = 4.5√3 m/s * 0.76 seconds - (1/2) * 9.8 m/s^2 * (0.76 seconds)^2
Simplifying the equation, we can find the value of h:
h ≈ 1.93 meters
Therefore, the maximum height the basketball will reach is approximately 1.93 meters.