A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100 m/s. The rocket moves for 3.00 s a long its initial line of motion with an acceleration of 30.0 m/s/s. At this time, its engines fail and the rocket proceeds to move as a projectile. Find:&nbsp;a) the maximum altitude reached by the rocket&nbsp;b) its total time of flight&nbsp;c) its horizontal range.

Question

A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100 m/s. The rocket moves for 3.00 s a long its initial line of motion with an acceleration of 30.0 m/s/s. At this time, its engines fail and the rocket proceeds to move as a projectile. Find:&amp;nbsp;a) the maximum altitude reached by the rocket&amp;nbsp;b) its total time of flight&amp;nbsp;c) its horizontal range.

oppturf · Accepted Answer

Lets find the maximum altitude of the rocketLets not worry about the components until the rocket has done accelerating.The angle will remain constant during acceleration.The velocity the second the rocket stops accelerating isVf=v+atVf=100+30(3)Vf=190m/s still acting at 53 degrees.how far has it moved in each directiond=Vb+12d=100(3)+15(9)d=435mLets look ar drawingSo we can do some basic trig to find its height and horizontal distancesin=opphyp,&amp;nbsp;cos⁡53=distance435,&amp;nbsp;distance=357.4&amp;nbsp;msin=adjhyp,&amp;nbsp;cos⁡53=distance435,&amp;nbsp;distance=261.7&amp;nbsp;mnow we have the height and distance when it stops accelerating. Now its a simple projectile motion questionGoing back we found it had a velocity of 190 m/sLets split that into x and y componentVelocity&amp;nbsp;y=sin⁡53⋅190=151.7&amp;nbsp;m/sVelocity&amp;nbsp;x=cos⁡53⋅190=114.3&amp;nbsp;m/sa) Max altitude just concerned with y component now our old Vf is our new Vi when it reaches its maximum height the velocity final will be zero since gravity is accelerating it downVf=V+at0=151.7-9.8t&amp;nbsp;t=−151.7−9.8&amp;nbsp;t=15.4sAgaind=Vb+12atd=151.7(15.4)−4.9(15.4)2d=2336m-1162m=1174But that was after the engines remember before the engines blew it had a height of 357.4 mso total height =1174+357.4=1531.6 mb) so right now we know the total time to its max height but we still need to find the time it takes to free fall to earthd=Vb+12atV1=0&amp;nbsp;nowd=4.9t2t2=1531.64.9=312.6t=17.6 sso lets add up the total times3 s till engine fall , 15.4 s till it reaches max alt and 17.6 seconds downTotal time = 36 sc) horinzontal rangeonce the engines blew the x velocity was constant.recallVelocity&amp;nbsp;x=cdos53⋅190=114.3&amp;nbsp;m/sso d=td=114.3(33) the first 3 seconds werent constantd=3771.9but we must add on the original distance when engine blewdistance =261.7mso total distance =3771.9+261.7=4033.6 m

Jeffrey Jordon · Accepted Answer

SolutionSpeed of the rocket after 3seconds =100+3×3=190m/sdisplacement =&amp;nbsp;100×3+12×30×32=435mvertical displacement =&amp;nbsp;435sin⁡53∘=348mhorizontal displacement =&amp;nbsp;435cos⁡53∘=261max=oi^,&amp;nbsp;&amp;nbsp;ux=114i^ay=−gj^,&amp;nbsp;&amp;nbsp;uy=152j^H=uy22g=649.8mmax Altitude = 649.8 + 348 = 997.8 mtime of flight under gravity is give by−348=152T−12gT2⇒T=32.5stotal time of flight = 32.5 + 3 =35.5 shorizontal range =&amp;nbsp;114×32.5+261=3966m

nick1337 · Accepted Answer

Step 1:a) To find the maximum altitude reached by the rocket, we need to determine the vertical component of its velocity at that point.Let&#039;s start by calculating the initial vertical velocity component (v0y) using the initial speed (v0) and the launch angle (θ):v0y=v0·sin(θ)Given that v0=100 m/s and θ=53∘, we can substitute these values to find v0y:v0y=100·sin(53∘)Next, we can calculate the time taken (t) to reach the maximum altitude using the following formula:t=v0ygwhere g is the acceleration due to gravity. Since the rocket moves as a projectile after the engines fail, we assume g=9.8 m/s2.Substituting the values, we have:t=100·sin(53∘)9.8Finally, the maximum altitude (h{max}) can be calculated using the formula:h{max}=v0y·t−12·g·t2Substituting the known values, we get:h{max}=100·sin(53∘)·100·sin(53∘)9.8−12·9.8·(100·sin(53∘)9.8)2Now, let&#039;s calculate the value for h{max}:h{max}=100·sin(53∘)·100·sin(53∘)9.8−12·9.8·(100·sin(53∘)9.8)2Step 2:b) To find the total time of flight (T), we need to consider the time taken for the rocket to reach its maximum altitude (t) and the time taken for it to descend from the maximum altitude back to the ground.Since the rocket moves as a projectile after the engines fail, its total time of flight is twice the time taken to reach the maximum altitude (t).Therefore, T=2·t.Substituting the value of t that we calculated earlier, we have:T=2·100·sin(53∘)9.8Now, let&#039;s calculate the value for T:T=2·100·sin(53∘)9.8Step 3:c) To find the horizontal range (R) of the rocket, we need to determine the horizontal component of its velocity (vx) and the total time of flight (T).The horizontal component of velocity (vx) remains constant throughout themotion since no horizontal force acts on the rocket after the engines fail. Therefore, we can calculate vx using the initial speed (v0) and the launch angle (θ):vx=v0·cos(θ)Given v0=100 m/s and θ=53∘, we can substitute these values to find vx:vx=100·cos(53∘)Finally, the horizontal range (R) can be calculated using the formula:R=vx·TSubstituting the known values, we get:R=100·cos(53∘)·2·100·sin(53∘)9.8Now, let&#039;s calculate the value for R:R=100·cos(53∘)·2·100·sin(53∘)9.8

RizerMix · Accepted Answer

a) To find the maximum altitude reached by the rocket, we can use the kinematic equation for vertical motion. The formula for the maximum height (h) reached by a projectile launched with an initial velocity (v₀) at an angle (θ) with respect to the horizontal is given by:h=v₀2sin2(θ)2gwhere v₀ is the initial velocity of the rocket, θ is the launch angle, and g is the acceleration due to gravity. Substituting the given values into the formula:h=1002sin2(53∘)2·9.8Calculating this expression will give us the maximum altitude reached by the rocket.b) The total time of flight can be determined using the equation:T=2v₀sin(θ)gwhere T represents the total time of flight. Plugging in the known values:T=2·100sin(53∘)9.8Evaluating this expression will give us the total time of flight.c) The horizontal range, denoted as R, can be calculated using the formula:R=v₀cos(θ)·twhere R represents the horizontal range and t is the time of flight. Substituting the given values:R=100cos(53∘)·3.00

Don Sumner · Accepted Answer

To solve the given problem, we can break it down into three parts:a) Maximum altitude reached by the rocket:We can use the kinematic equation for vertical motion to find the maximum altitude (hmax) reached by the rocket. The initial vertical velocity (vy0) can be found using the initial speed and launch angle:vy0=v0sin(θ)where v0 is the initial speed of the rocket and θ is the launch angle.The time taken for the rocket to reach its maximum altitude can be found using the equation:tmax=vy0gwhere g is the acceleration due to gravity.Substituting the values into the equations, we get:hmax=vy0·tmax−12g·tmax2b) Total time of flight:The total time of flight (T) can be found by summing up the time taken for the rocket to reach its maximum altitude and the time taken for it to fall back to the ground. Since the rocket moves for 3.00 s with an acceleration of 30.0 m/s/s, we can calculate the time taken to reach the maximum altitude using:tmax=3.00sTo find the time taken for the rocket to fall back to the ground, we can use the equation for vertical motion:h=vy0·t−12g·t2Setting h to zero and solving for t gives us the time taken for the rocket to fall back to the ground.c) Horizontal range:The horizontal range (R) can be calculated using the equation:R=vx0·Twhere vx0 is the initial horizontal velocity of the rocket and T is the total time of flight.

A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100 m/s. The rocket moves for 3.00 s a long its initial l

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