The height above ground of a rock dropped from a bridge can be found using the formula h(t) = 9.8tsquared + 100, where t represents the number of seconds.
What is the height of the rock after 3 seconds?
How long will it take the rock to hit the bottom?
Your equation should read
h(t) = -9.8t^2 + 100
when t = 3
h(3) = -9.8(9) + 100
= 11.8
it will be 11.8 m above the ground
how long to hit ground ??
when h(t) = 0
0 = -9.8t^2 + 100
9.8t^2 = 100
t^2 = 100/9.8 = 10.204..
t = √10.204.. = appr 3.2 seconds
To find the height of the rock after 3 seconds, we can substitute t = 3 into the formula h(t) = 9.8t^2 + 100 and solve for h(3).
h(3) = 9.8(3)^2 + 100
= 9.8(9) + 100
= 88.2 + 100
= 188.2
Therefore, the height of the rock after 3 seconds is 188.2 units above the ground.
To find out how long it will take for the rock to hit the bottom, we need to find the value of t when h(t) = 0. This is because when the rock hits the bottom, its height will be zero.
0 = 9.8t^2 + 100
To solve this equation, we can move 100 to the other side:
9.8t^2 = -100
Now, divide both sides of the equation by 9.8:
t^2 = -100/9.8
Taking the square root of both sides, we get:
t = ±√(-100/9.8)
Since the square root of a negative number is not a real number, the equation has no solution in terms of time. This means that the rock will never hit the bottom, according to the given model.