What is the Maximum Height a Basketball Reaches When Thrown Upward?

[swirly90]

A referee throws a basketball veritcally upward with an initial speed of 5.0m/s. Determine the maximum height above the floor reached by the basketball if it starts from a height 1.5m


How do you solve this problem? I have tried but I just don't seem to have enough information. I know acceleration due to gravity is 9.8m/s. I need at least time, initial velocity and acceleration to solve this but I really have no clue how to solve this.

The asnwer is 2.8m

 

[mikelepore]

An object thrown upward stops for an instant before it falls back down, so you can make a statement about the final velocity of the upward part of the motion.

 

[tomcue]

To solve this problem, we can use the equations of motion for a vertically thrown object. The initial height of the basketball is given as 1.5m and the initial velocity is 5.0m/s. We also know that the acceleration due to gravity is -9.8m/s^2, as it acts in the opposite direction of the initial velocity.

To determine the maximum height reached, we can use the equation:

h = h0 + v0t + 1/2at^2

Where h is the final height, h0 is the initial height, v0 is the initial velocity, a is the acceleration, and t is the time.

In this case, our final height is the maximum height reached, h0 is 1.5m, v0 is 5.0m/s, and a is -9.8m/s^2. We can rearrange the equation to solve for t, which is the time it takes for the basketball to reach its maximum height.

t = (v0 - √(v0^2 + 2ah0)) / a

Plugging in our values, we get:
t = (5.0 - √(5.0^2 + 2(-9.8)(1.5))) / -9.8
t = (5.0 - √(25 + (-29.4))) / -9.8
t = (5.0 - √(-4.4)) / -9.8
t = (5.0 - 2.1) / -9.8
t = 2.9 / -9.8
t = -0.3 seconds

Since time cannot be negative, we know that the basketball reaches its maximum height after 0.3 seconds.

To determine the maximum height reached, we can plug this time back into our original equation:

h = 1.5 + (5.0)(0.3) + 1/2(-9.8)(0.3)^2
h = 1.5 + 1.5 - 0.44
h = 2.8m

Therefore, the maximum height reached by the basketball is 2.8m above the floor.