wo identical rubber balls are dropped from different heights. Ball 1 is dropped from a height of 100 feet, and ball 2 is dropped from a height of 210 feet. Write a function for the height of each ball.
h1(t) =
h2(t) =
When does ball 1 reach the ground? Round to the nearest hundredth. _____seconds
When does ball 2 reach the ground? Round to the nearest hundredth. _____seconds
h1(t) = -16t^2 + 100
h2(t) = -16t^2 + 210
when ball 1 reaches ground, h1(t) = 0
16t^2 = 100
t^2 = 100/16
t = √(100/16) = 10/4 = 5/2 seconds
do the 2nd the same way
Had no idea that Lisa had asked this question several times before and had shown no sign of effort.
(I started answering top down)
Oh well.
If her guile and persistence have helped her learn the material, it has worked out ok.
To write the equations for the height of each ball over time, we can use the equation of motion for freely falling objects. The equation is:
h(t) = h0 - 0.5 * g * t^2
Where:
- h(t) is the height of the object at time t
- h0 is the initial height from which the object is dropped
- g is the acceleration due to gravity (approximately 32.2 feet per second squared)
- t is the time in seconds
For ball 1, h1(t), the equation will be:
h1(t) = 100 - 0.5 * 32.2 * t^2
For ball 2, h2(t), the equation will be:
h2(t) = 210 - 0.5 * 32.2 * t^2
Now, let's calculate when each ball reaches the ground by setting h(t) to 0 and solving for t.
For ball 1:
0 = 100 - 0.5 * 32.2 * t^2
Simplifying the equation:
0.5 * 32.2 * t^2 = 100
Dividing both sides by 0.5 * 32.2:
t^2 = 100 / (0.5 * 32.2)
t^2 ≈ 3.105590
Taking the square root of both sides, we get:
t ≈ √3.105590
t ≈ 1.76 seconds
Rounding to the nearest hundredth, ball 1 reaches the ground in approximately 1.76 seconds.
For ball 2:
0 = 210 - 0.5 * 32.2 * t^2
Simplifying the equation:
0.5 * 32.2 * t^2 = 210
Dividing both sides by 0.5 * 32.2:
t^2 = 210 / (0.5 * 32.2)
t^2 ≈ 6.510563
Taking the square root of both sides, we get:
t ≈ √6.510563
t ≈ 2.55 seconds
Rounding to the nearest hundredth, ball 2 reaches the ground in approximately 2.55 seconds.