NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)=−4.9t^2+115t+220.
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after ? seconds.
How high above sea-level does the rocket get at its peak?
The rocket peaks at ? meters above sea-level.
splashdown occurs when h=0
max height is at t = -b/2a = 115/9.8
To find the time of splashdown, we need to find the time when the rocket's height above sea-level is equal to 0. So, we can set the equation h(t) = -4.9t^2 + 115t + 220 equal to 0 and solve for t.
-4.9t^2 + 115t + 220 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -4.9, b = 115, and c = 220. Plugging in these values, we get:
t = (-115 ± √(115^2 - 4*(-4.9)*(220))) / (2*(-4.9))
Now, we can calculate the values under the square root and simplify the equation further.
t = (-115 ± √(13225 + 4312)) / (-9.8)
t = (-115 ± √17537) / (-9.8)
Now, we can calculate the two possible values for t by evaluating both the positive and negative square root.
t ≈ (-115 + 132.43) / (-9.8) or t ≈ (-115 - 132.43) / (-9.8)
t ≈ 17.43 / (-9.8) or t ≈ -247.43 / (-9.8)
Since time cannot be negative in this context, we ignore the negative value. Thus,
t ≈ 17.43 / (-9.8) ≈ -1.78 seconds
However, this negative value does not make physical sense, so we discard it. The time of splashdown is approximately
t ≈ 17.5 seconds.
To find the height at the peak, we can determine the maximum value of the equation h(t). As the rocket reaches its peak, the velocity becomes zero, indicating the highest point.
To find this maximum point, we can identify the t-value at which the derivative of h(t) is equal to zero. The derivative of h(t) is obtained by differentiating the function with respect to t:
h'(t) = -9.8t + 115
Setting h'(t) = 0, we can solve for t:
-9.8t + 115 = 0
-9.8t = -115
t ≈ -115 / -9.8
t ≈ 11.73 seconds
We can now substitute this value of t into the original equation to find the height at the peak:
h(t) = -4.9t^2 + 115t + 220
h(11.73) ≈ -4.9(11.73)^2 + 115(11.73) + 220
h(11.73) ≈ -667.989 + 1349.95 + 220
h(11.73) ≈ 902.961
Therefore, the rocket reaches a peak height of approximately 902.961 meters above sea-level.