If a rock is thrown upward on the planet Mars with a velocity of 19 m/s, its height (in meters) after t seconds is given by H = 19t − 1.86t^2. When will the rock hit the surface?
well, what is t when h=0?
h(t) = t(19-1.86t)
I misread the question!! I was thinking that setting it equal to 0 would be something completely different. Thank you so much!
To find when the rock will hit the surface, we need to find the time at which its height is equal to zero. In other words, we need to solve the equation H = 0.
Given the equation H = 19t − 1.86t^2, we can substitute H with 0 and solve for t:
0 = 19t − 1.86t^2
We can rewrite the equation in the form of a quadratic equation:
1.86t^2 - 19t = 0
To solve this equation, we can factor out t:
t(1.86t - 19) = 0
Setting each factor to zero:
t = 0 or 1.86t - 19 = 0
From the first factor, t = 0 represents the initial position of the rock when it was thrown. We are interested in when it will hit the surface, so t = 0 is not the solution we are looking for.
Now, let's solve the second factor:
1.86t - 19 = 0
Adding 19 to both sides:
1.86t = 19
Dividing both sides by 1.86:
t = 19 / 1.86
Calculating this value, we find:
t ≈ 10.22 seconds
Therefore, the rock will hit the surface of Mars approximately 10.22 seconds after it was thrown upward.