A ball is thrown upward from the ground with an initial speed of 23.2 m/s; at the same instant, another ball is dropped from a building 20 m high. After how long will the balls be at the same height?
let up be +
ball first:
h=23.2*t-4.9t^2
ball second
h=20-4.9t^2 set the two equations equal, solve for t. Notice is is a pretty simple solution.
To find out when the two balls will be at the same height, we can set up equations for the heights of each ball as a function of time.
Let's first find the equation for the ball thrown upward. We know that the initial speed of the ball is 23.2 m/s, and the acceleration due to gravity is -9.8 m/s² (negative because it acts in the opposite direction of the ball's motion).
The equation for the height of the ball thrown upward as a function of time can be described by the formula:
h₁(t) = v₀t + (1/2)at²
Where:
h₁(t) is the height of the ball thrown upward at time 't'
v₀ is the initial velocity (23.2 m/s)
a is the acceleration due to gravity (-9.8 m/s²)
t is the time
Next, let's find the equation for the ball dropped from the building. Since it is dropped, its initial speed is 0 m/s. The only force acting on it is gravity, so it falls with a constant acceleration due to gravity.
The equation for the height of the ball dropped from the building as a function of time can be described by the formula:
h₂(t) = h₀ + v₀t + (1/2)at²
Where:
h₂(t) is the height of the ball dropped from the building at time 't'
h₀ is the initial height (20 m)
v₀ is the initial velocity (0 m/s)
a is the acceleration due to gravity (-9.8 m/s²)
t is the time
Now we can find when the two balls will be at the same height by equating the two height equations and solving for 't':
h₁(t) = h₂(t)
v₀t + (1/2)at² = h₀ + v₀t + (1/2)at²
Simplifying the equation:
v₀t - v₀t + (1/2)at² - (1/2)at² = h₀
Now we can cancel out the like terms:
0 = h₀
This tells us that when both balls are at the same height, the height will be zero. Therefore, the balls will be at the same height when they are on the ground.