### Rocket Launch ProblemA rocketry club is holding a competition. There is cloud cover at 1000 feet. If a rocket is launched with an initial upward velocity of 315 ft/sec, use the function:\[ h(t) = -16t^2 + vt + s \]to determine how long the rocket is out of sight.### ExplanationThe function \( h(t) = -16t^2 + vt + s \) is a mathematical model used to calculate the height \( h \) of the rocket over time \( t \). In this equation:- \( -16t^2 \) represents the effect of gravity (assuming \( t \) is in seconds and \( h \) in feet).- \( v \) is the initial velocity of the rocket, given as 315 ft/sec.- \( s \) is the initial position of the rocket, typically the launchpad height, assumed here to be 0 feet.#### ObjectiveThe goal is to find the duration for which the rocket remains above the cloud cover (1000 feet), making it out of sight.### Calculations & GraphsTo solve the problem, graph the function to find the time interval during which the height \( h(t) \) is above 1000 feet. Initially, set \( h(t) = 1000 \) and solve for \( t \):\[ 1000 = -16t^2 + 315t \]By solving this quadratic equation, the values of \( t \) where the rocket crosses 1000 feet will provide the required time interval. Use these roots to find when the rocket passes above and then below 1000 feet, marking the "out of sight" duration.

The Velocity of a body is the first time derivative of the distance travelled by the body. Mathematically, v=dxdt where, x is the distance travelled by the moving body t is the time taken by the bodyThe function of height gained by the rocket is given as: h(t)=-16t2+vt+s First of applying the boundary condition, at t=0sec ; h(t)=0 feet , we get 0=-16×02+(v×0)+ss=0 thus, the magnitude of the variable s came out to be zero. Now, the function becomes, h(t)=-16t2+vt...............................(a) now, differentiating the above function with respect to t once, we get dh(t)dt=-16dt2dt+ddtvt+ddt0v(t)=-32t+v..............................................(1) Now, applying the boundary conditions in equation (1) as: at t=0 ; v(0)=315 ftsec, we get 315=-32×0+v315=0+vv=315 fts Thus, the magnitude of v came out to be 315 ft/sec. Now applying another boundary condition in equation (1) as: At t=t sec ; v(t)=0, we get v(t)=-32t+v0=-32t+31532t=315t=31532t=9.84 seconds Thus, the time taken by the rocket to reach the topmost point of its journey be 9.84 seconds. Since, it is given in the question that the rocket is projected vertically upwards with a velocity of 315 ft/sec, thus calculating the time it would take to reach 1000 feets from the ground. putting h(t)=1000feet in equatin (a), and calculating t as: h(t)=-16t2+vt1000=-16t2+315t16t2-315t+1000=0 After solving the above quadratic equation, we get t=3.97 sec Thus, the time taken by the rocket to reach 1000 feet from ground is 3.97 seconds. now, we know that the time taken by the rocket to reach its apex point of journey is 9.84 seconds, but it covers 1000 feets in just 3.97 seconds, thus for the remaining time the rocket was out of sight. Thus, to=9.84-3.97=5.87 seconds Thus, in the first part of its journey the rocket remains out of sight for 5.87 seconds. Now for both the parts of the journey (going up and coming down), it will remain out of sight for twice the time. Thus, To=2to=2×5.87=11.74 sec The Total time for which the rocker remain out of sight is 11.74 seconds.

Engineering

Mechanical Engineering

A rocketry club is holding a competition. There is cloud cover at 1000 feet. If a rocket is launched with an initial upward velocity of 315 ft/sec, use the function h(1)=-161 +vt+s to determine how long the rocket is out of sight.

A rocketry club is holding a competition. There is cloud cover at 1000 feet. If a rocket is launched with an initial upward velocity of 315 ft/sec, use the function h(1)=-161 +vt+s to determine how long the rocket is out of sight.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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