A volcano launches a lava bomb straight upward with an initialspeed of 23 m/s. T

Question

A volcano launches a lava bomb straight upward with an initialspeed of 23 m/s. Taking upward to be thepositive direction, find the speed and direction of the lava bombafter the following amounts of time have elapsed.' a) 1.6s speed? velocity? b) 3.8s speed? velocity? pleez explain it completely with the work and answer! thanks!I promise to award live-saver!! A volcano launches a lava bomb straight upward with an initialspeed of 23 m/s. Taking upward to be thepositive direction, find the speed and direction of the lava bombafter the following amounts of time have elapsed.' a) 1.6s speed? velocity? b) 3.8s speed? velocity? pleez explain it completely with the work and answer! thanks!I promise to award live-saver!!

Explanation / Answer

intial speed u= 23m/s t = 1.6s from the kinematic eq = s = ut - 1/2at2 plug in values s = -------m when t = 1.6s speed = s/t velocity = v = s/t plug in values and do caliculations ---------------------------------------------------------- intial speed u= 23m/s t = 3.8s from the kinematic eq displacement is s = ut -(1/2)gt2 plug in values s = -------m velocity = v = s/t Now we find  total distance travelled isS time taken to reach maximumheight is t1 = u /g                                                                 = -------- sec Now the remaining time is t2 = t - t1 =---------sec The distance travelled in a time t2 from the maximumheight                                                   is S = ut + (1/2)gt2                                                                =0 + (1/2)gt2                                                                = -------- m     Then the total distance travelled intime t = 3.8 s is D = H + S                                                                                        =--------- m      Then the speed (intime 3.8s ) = D /3.8s                                                    =-------- m/s                         plug in values s = -------m when t = 1.6s speed = s/t velocity = v = s/t plug in values and do caliculations ---------------------------------------------------------- intial speed u= 23m/s t = 3.8s from the kinematic eq displacement is s = ut -(1/2)gt2 plug in values s = -------m velocity = v = s/t Now we find  total distance travelled isS time taken to reach maximumheight is t1 = u /g                                                                 = -------- sec Now the remaining time is t2 = t - t1 =---------sec The distance travelled in a time t2 from the maximumheight                                                   is S = ut + (1/2)gt2                                                                =0 + (1/2)gt2                                                                = -------- m     Then the total distance travelled intime t = 3.8 s is D = H + S                                                                                        =--------- m      Then the speed (intime 3.8s ) = D /3.8s                                                    =-------- m/s                         t = 3.8s from the kinematic eq displacement is s = ut -(1/2)gt2 plug in values s = -------m velocity = v = s/t Now we find  total distance travelled isS time taken to reach maximumheight is t1 = u /g                                                                 = -------- sec Now the remaining time is t2 = t - t1 =---------sec The distance travelled in a time t2 from the maximumheight                                                   is S = ut + (1/2)gt2                                                                =0 + (1/2)gt2                                                                = -------- m     Then the total distance travelled intime t = 3.8 s is D = H + S                                                                                        =--------- m      Then the speed (intime 3.8s ) = D /3.8s                                                    =-------- m/s                         plug in values s = -------m velocity = v = s/t Now we find  total distance travelled isS time taken to reach maximumheight is t1 = u /g                                                                 = -------- sec Now the remaining time is t2 = t - t1 =---------sec The distance travelled in a time t2 from the maximumheight                                                   is S = ut + (1/2)gt2                                                                =0 + (1/2)gt2                                                                = -------- m     Then the total distance travelled intime t = 3.8 s is D = H + S                                                                                        =--------- m      Then the speed (intime 3.8s ) = D /3.8s                                                    =-------- m/s                        

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