Height of a toy rocket. A toy rocket is launched straight up from the top of abu

Question

Height of a toy rocket. A toy rocket is launched straight up from the top of abuilding 50 ft tall at an initial velocity of 200 ft per sec. a) Give the function that describes the height of the rocketin terms of time t. b) Determine the time at which the rocket reaches its maximumheight, and the maximum height in feet. c) For what time interval will the rocket be more than 300 ftabove ground level? d) After how many seconds will it hit the ground? Thanks much, Height of a toy rocket. A toy rocket is launched straight up from the top of abuilding 50 ft tall at an initial velocity of 200 ft per sec. a) Give the function that describes the height of the rocketin terms of time t. b) Determine the time at which the rocket reaches its maximumheight, and the maximum height in feet. c) For what time interval will the rocket be more than 300 ftabove ground level? d) After how many seconds will it hit the ground? Thanks much, b) Determine the time at which the rocket reaches its maximumheight, and the maximum height in feet. c) For what time interval will the rocket be more than 300 ftabove ground level? d) After how many seconds will it hit the ground? Thanks much,

Explanation / Answer

a)the reqd fuction is h(t)=ut-gt2/2   h=height g=acceleration due to gravity t=time u=initial velocity b)now at highest point velocity is zero we can use the formula v=u-gt where v=finalvelocity                                    0=200-32t g=32ft/sec2                                     t=200/32=6.25 so the roket will reach its highest point in 6.25 sec now putting t=6.25 h(6.25)=200*6.25-32*6.252/2            =625 max ht is 625ft above buildin or 625+50=675 ft fromground c)now to be 300 ft above ground the roket must be 250 ft abovebuilding    putting h(t)=250    250=200t-16t2     16t2-200t+250=0      8t2-100t+125=0 dividin by 4     solve for t ...there will be 2 valuesof t since roket passes 300 ft twice once during ascent and duringdescent..thir difference will give ur reqd time interval d)during descent let it take t secs to reach ground h(t)=ut+gt2/2 positive sign for it is goingdown 675=0t+16t2 since u=0 at highest point t=(675/16) t=6.49 sec the roket will reach the ground after(6.49+6.25)secs=12.74secs after launch t=(675/16) t=6.49 sec the roket will reach the ground after(6.49+6.25)secs=12.74secs after launch

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