A toy rocket is launched from the top of a building 106 feet...

A toy rocket is launched from the top of a building 106 feet tall at an initial velocity of 194 feet per second. a) Give the function that describes the height of the rocket in terms of time t. b) Determine the time at which the rocket reaches it maximum height, and the maximum height in feet. c) For what time interval will the rocket be more than 302 feet above ground level? d) After how many seconds will it hit the ground? a) The function that describes the height of the rocket in terms of time t is s(t) b) The rocket reaches it maximum height of _______ feet after approximately ________ seconds. (Round to the nearest hundredth as needed.) c) The rocket will be more than 302 feet above ground level for all values of t in the interval ________. (Type your answer in interval notation. Do not round until the final answer. Then round ro the nearest hundredth) d) The rocket will hit the ground after _____ seconds. (Round to the nearest hundredth as needed.)

Transcript

00:01 All right, so we have this model rocket.

00:02 It's launched from a platform.

00:04 Initial velocity is 194 feet per second.

00:07 So the vertical motion model formula, h of t, the height in relation to time, is negative 16t squared plus v -0t plus h -0.

00:23 Okay, so v -0 is the initial velocity.

00:26 H -0 is the initial.

00:30 Velocity or h not so this is a little height so plugging our numbers in h of t equals negative 16 t squared v not the initial velocity velocity is 194 feet per second we threw it up so plus 194 times t and then our initial height is the height of the platform which is 106 feet this is our function in terms of t, height in terms of time.

01:07 Okay, now to determine the time the rocket reaches the maximum height as well as the maximum height, we need to find the vertex.

01:16 So if you're in pre -calculus or algebra 2, you do this.

01:22 You do negative b over 2a to get the x coordinate of the vertex.

01:27 That's what x is going to equal.

01:30 And then you plug it back in to find the y core.

01:33 If you're in calculus, you take the derivative and set the derivative equal to 0.

01:39 So the derivative is h prime of t equals negative 32 t plus 194, and then we would set this equal to 0.

01:55 Subtract 194, divide by 32.

01:59 That's going to come out to be 6.

02:06 Something.

02:10 So 194 divided by 32, 6 .0625.

02:15 So when you solve this, t is going to come out to be 6 .025, which is that's the time that it reaches the maximum height.

02:28 That's if you're in calculus.

02:30 If you're in algebra 2, a pre -calculus, and you haven't no derivatives yet, you do this.

02:36 So you'd get 1904.

02:38 Divided by 2 times negative 16, which is still going to give you 6 .025.

02:47 Okay? so our max time, our time, it's going to reach the maximum height at 6 .025 seconds.

02:58 Okay? to find the maximum height, you just take that 6 .025 and plug it into our function from part a.

03:12 So 6 .025 square it times it by negative 16 and then we're going to add 194 times 6 .025 and then we're going to add 106.

03:30 So 694 .04 feet is the maximum height that we get...

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